AAT SOLUTIONS - CH13.pdf

Chapter 13 Prerequisite Skills (p. 850)

2.

f ( g(x)) 5 x and g ( f(x)) 5 x. of 1 unit is x 2 1 y 2 5 1.

5.

a2 1 b2 5 c2

8 2 1 10 2 5 c 2

(2.5)2 1 b 2 5 (6.5)2

64 1 100 5 c

2

6.25 1 b 2 5 42.25

164 5 c

2

2

}

Ï 4 Ï 41 5 c

hyp

17

adj

8

sec u 5 } 5} 8 adj cot u 5 } opp 5 } 15

u 5

5

Ï2 5 1 Ï2 5Ï2 } adj Ï2 5 1 } 5 cos u 5 } } 5 } } 5 } 2 hyp Ï2 5Ï 2 opp 5 5 }5 5 1 tan u 5 } adj } hyp } 5Ï 2 5 Ï2 csc u 5 } opp 5 } 5 } hyp } 5Ï2 } 5 Ï2 5 sec u 5 } 5 adj adj 5 51 cot u 5 } opp 5 } 5

sin u 5 } 5} } 5 } } 5 } 2 hyp

7. 4 x 2 5 5 3

6. b 5 9, c 5 11 2

4x 5 8

a 2 1 9 2 5 112

2 2x 5 23

2

2x 5 3

a 1 81 5 121

3

a 2 5 40

}

opp

c 5 2Ï41

x 5 }2 }

a 5 2Ï10 8. log2x 5 21

x 5 221 4.

1

x 5 }2 Copyright © by McDougal Littell, a division of Houghton Mifflin Company.

15

17

b56 }

a 1b 5c

opp

5 2

2Ï41 5 c

2

8

3.

}

2

adj

hyp

csc u 5 } opp 5 } 15

5} tan u 5 } 8 adj

b 5 36

}

15

5} cos u 5 } 17 hyp

a 5 2.5, c 5 6.5

a2 1 b2 5 c2

opp

sin u 5 } 5} 17 hyp

3. An equation of the circle with center (0, 0) and a radius

a 5 8, b 5 10

8

15

2. Functions f and g are inverses of each other if

4.

u

17

4 5 1. The reciprocal of } is }. 5 4

10

9. 25 1 2ln 3x 5 20

10. x 1 65 1 55 5 180

2ln 3x 5 25

x 1 120 5 180

ln 3x 5 12.5

u 7

x 5 60

2

7 1 x 2 5 10 2

12.5

3x 5 e

x 2 5 51 }

3x ø 268,337.29

x 5 Ï51

x ø 89,446

}

opp

11. x 1 x 1 90 5 180

12. x 1 41 1 107 5 180

2x 1 90 5 180

x 1 148 5 180

2x 5 90

5.

B

x 5 32

Lesson 13.1

Ï51

sin u 5 } 5} 10 hyp

458

5

x 5 45 A

a

b

C

A 5 908 2 458 5 458

13.1 Guided Practice (pp. 853–855)

b

1. 5

x

3

u

cos 458 5 }5

5(sin 458) 5 b

5(cos 458) 5 a

}

4

opp

3

adj

4

opp

3

sin u 5 } 5 }5 hyp 5 }5 cos u 5 } hyp 5 }4 tan u 5 } adj

hyp

5

hyp

5

adj

4

csc u 5 } opp 5 } 3 sec u 5 } 5 }4 adj

a

sin 458 5 }5

}

1 2

2 5 } 5a 2

5Ï2 }5b

}

5Ï2 }5a

3.54 ø b

3.54 ø a

Ï2 5 } 5b 2

2

1Ï 2 }

2

cot u 5 } opp 5 } 3

Algebra 2 Worked-Out Solution Key

n2ws13-a.indd 713

713

6/27/06 11:39:00 AM

Chapter 13, 6.

continued 10.

B c

C

h

sin 388 5 } 300

B 5 908 2 328 5 588 10

a

tan 328 5 } 10

cos 328 5 } c

10(tan 328) 5 a

c(cos 328) 5 10 10 cos 328

6.25 ø a 7.

300 sin 388 5 h 184.7 ø h The height of the parasailer is about 184.7 feet.

c5}

13.1 Exercises (pp. 856 – 858)

c ø 11.79

Skill Practice

B

20

h

388

328 10

A

300 ft

a

1. An angle of elevation is the angle formed by the line

of sight to an object above you and a line parallel to the ground.

a

2. Solving a right triangle means to find all the unknown

side lengths and angle measures of the triangle. 718 A b C

}

3. x 5 Ï 10 2 1 24 2 5 26

B 5 908 2 718 5 198 a sin 718 5 } 20

b cos 718 5 } 20

20(sin 718) 5 a

20(cos 718) 5 b

18.91 ø a

6.51 ø b

B c A

608

b

24

opp

24

12

adj

10

5

opp

24

12

sin u 5 } 5} 5} 26 13 hyp

}

13

hyp

26

13

adj

10

5

hyp

Ï73

hyp

Ï73

adj

8

cot u 5 } 5} opp 5 } 24 12

4. x 5 Ï 8 2 1 3 2 5 Ï 73

A 5 908 2 608 5 308 7

b

tan 608 5 }7

cos 608 5 }c 7 cos 608

7 tan 608 5 b

73

x5

7 c5} 0.5

7Ï 3 5 b 12.12 ø b

c 5 14 Widforss Point

3

8 }

opp

3 3Ï73 Ï73 } adj 8 8Ï73 } 5 cos u 5 } } 5 } 73 hyp Ï73 opp 3 5 }8 tan u 5 } adj

sin u 5 } 5} } 5 } 73 hyp

}

d

}

u

c5}

}

Powell Point 768 2 mi

26

sec u 5 } 5} 5} 5 10 adj

5} 5} tan u 5 } 5 10 adj

C

hyp

csc u 5 } 5} opp 5 } 24 12

5} 5} cos u 5 } 26 13 hyp

7

9.

x 5 26

u

}

csc u 5 } opp 5 } 3 }

sec u 5 } 5} 8 adj cot u 5 } opp 5 } 3

5. x 5 Ï 112 2 8 2 5 Ï 57

}

u x5 Yavapai Point

2

cos 768 5 }d d cos 768 5 2 2 cos 768

d5} d ø 8.27

57

11

8

hyp opp 8 11 sin u 5 } 5} csc u 5 } opp 5 } 11 8 hyp } } adj hyp Ï 57 11Ï57 11 } }5} 5 sec u 5 cos u 5 } } 5 } 57 11 hyp adj Ï 57 } } adj opp Ï 57 8 8Ï57 5} tan u 5 } } 5 } cot u 5 } opp 5 } 57 8 adj Ï 57

Copyright © by McDougal Littell, a division of Houghton Mifflin Company.

8.

10

The distance is about 8.27 miles.

714


n2ws13-a.indd 714

6/27/06 11:39:11 AM

Chapter 13, }

continued }

6. x 5 Ï 15 2 2 7 2 5 4Ï 11

9. x 5 Ï 6 2 2 5 2 5 Ï 11

}

x 5 4 11

7

u

u

6

15

opp

hyp

7

sin u 5 } 5} 15 hyp

}

}

hyp 15 15Ï11 sec u 5 } 5} } 5 } 44 adj 4Ï11 }

adj

7Ï11

7 4Ï11

}

4Ï11

5} tan u 5 } } 5 } cot u 5 } opp 5 } 7 44 adj }

7. x 5 Ï 14 2 2 9 2 5 Ï 115

}

Ï11

opp

5 Ï 11

hyp

6

hyp

6 Ï11

adj

Ï 11

}

5Ï11

5} tan u 5 } } 5 } 11 adj

}

6Ï11

}

cot u 5 } opp 5 } 5

x 5 115 }

}

Ï115

10. x 5 Ï 8 2 2 5 2 5 Ï 39

sin u 5 } 5} 14 hyp adj

}

adj

cos u 5 } 5} 6 hyp

5} sec u 5 } } 5 } 11 adj

9

opp

x 5 11

csc u 5 } opp 5 } 5

u 14

5

15

csc u 5 } opp 5 } 7

adj 4Ï11 5} cos u 5 } 15 hyp opp

}

}

9

5} cos u 5 } 14 hyp

8

x 5 39

}

opp Ï115 5} tan u 5 } 9 adj

u 5

}

hyp 14Ï115 14 csc u 5 } } 5 } opp 5 } 115 Ï115 hyp

14

5} sec u 5 } 9 adj adj

9 Ï115

}

9Ï115


}} }

2 2Ï17

1 Ï17

}

Ï17

sin u 5 } 5} } 5 } } 5 } 17 hyp adj

8 2Ï17

opp

2

Ï39

hyp

8 Ï39

hyp

8

adj

5 Ï39

}

}

8Ï39

5 }5 sec u 5 } adj

2 17 u 8

opp

opp

csc u 5 } } 5 } opp 5 } 39

Ï(2Ï17 )2 2 82 5 2

x52

Ï39

5} tan u 5 } 5 adj

cot u 5 } } 5 } opp 5 } 115 8. x 5

}

opp

sin u 5 } 5} 8 hyp

4 Ï17

}

4Ï 17

}

5Ï39

cot u 5 } } 5 } opp 5 } 39 }

11. x 5 Ï 7 2 1 3 2 5 Ï 58

}

5} cos u 5 } } 5 } } 5 } 17 hyp 1

x 5 58

5 }8 5 }4 tan u 5 } adj hyp

}

2Ï17

}

5 Ï17 csc u 5 } opp 5 } 2 }

}

hyp Ï17 2Ï17 5} 5} sec u 5 } 8 4 adj adj

8

54 cot u 5 } opp 5 } 2

7

u 3 }

opp

7 Ï58

7Ï58

adj

3 Ï58

3Ï58

hyp

Ï58

hyp

Ï58

adj

3

sin u 5 } 5} } 5 } 58 hyp }

5} cos u 5 } } 5 } 58 hyp }

csc u 5 } opp 5 } 7 }

5} sec u 5 } 3 adj cot u 5 } opp 5 } 7


n2ws13-a.indd 715

715

6/27/06 11:39:20 AM

Chapter 13, }

continued

12. x 5 Ï 10 2 2 7 2 5 Ï 51

10

}

15. C;

}

x 5 Ï9 2 2 4 2 5 Ï65

}

7 9

x 5 65

u x 5 51

opp

u

7

4

sin u 5 } 5} 10 hyp

}

opp

Ï65

tan u 5 } 5} 4 adj

}

adj Ï51 5} cos u 5 } 10 hyp }

opp 7 7Ï51 5} tan u 5 } } 5 } 51 adj Ï51

1 1 16. The reciprocal of cosecant is } , not }. cos u sin u }

x 5 Ï112 2 7 2 5 6Ï2

}

hyp

10Ï51

10 Ï51 } adj Ï51 } 5 cot u 5 } opp 7

}

5} sec u 5 } } 5 } 51 adj 11

}

x56 2

u

13. x 5 Ï 12 2 2 5 2 5 Ï 119

}

7

hyp

11 6Ï2

}

11Ï2

csc u 5 } } 5 } opp 5 } 12 12

17.

x 5 119

8

608

y

u

x

opp

adj

tan 608 5 } adj

1 2

}

12 Ï119

18. }

12Ï119

csc u 5 } } 5 } opp 5 } 119

x

y 5 16 y

458

21

}

5Ï119

5 Ï119

cot u 5 } } 5 } opp 5 } 119

adj

opp

cos 458 5 } hyp

14. x 5 Ï 6 2 2 112 5 Ï 157

}

sin 458 5 } hyp

}

}

y Ï2 }5} } 2 Ï 21

x Ï2 }5} } 2 Ï 21 }

}

Ï 42 }5x

x 5 157

19.

308 x

y

6

11 Ï157

}

11Ï157

sin u 5 } 5} } 5 } 157 hyp adj

6 Ï157

}

6Ï157

5} cos u 5 } } 5 } 157 hyp opp

11

hyp

Ï157

hyp

Ï157

5} tan u 5 } 6 adj }

csc u 5 } opp 5 } 11

2

12

u

opp

Ï 42 }5y

2

11

8 y

}5}

8Ï 3 5 x

}

}

x

}

opp Ï119 5} tan u 5 } 5 adj

adj

cos 608 5 } hyp

Ï3 5 }8

5

5} cos u 5 } 12 hyp

hyp

adj

opp

}

Ï119

sin u 5 } 5} 12 hyp

opp

tan 308 5 } adj }

Ï3 3

x 12

} 5} }

4Ï 3 5 x

adj

cos 308 5 } hyp }

Ï3 2

12

}5} y }

y 5 8Ï3


5

}

5} sec u 5 } 6 adj

716


n2ws13-a.indd 716

6/27/06 11:39:32 AM

Chapter 13,

continued

20. A;

25. A

opp tan 308 5 } adj

608 x 8

}

Ï3 3

x 5

}5}

308 5

b

c

C

758 15 B

}

5Ï3 3

}5x

21. A

A 5 908 2 758 5 158 358 b

15

b

16

tan 758 5 } 15

cos 758 5 } c 15 cos 758

15(tan 758) 5 b C

a

B

55.98 ø b

B 5 908 2 358 5 558 a

b

cos 358 5 } 16

16(sin 358) 5 a

16(cos 358) 5 b

9.18 ø a

13.11 ø b

22. A

498

C

a

B

a

C

b

sin 498 5 } 27

c

12

B

27(cos 498) 5 b

20.38 ø a

17.71 ø b

27. A

A 5 908 2 538 5 378 b

32

12

tan 538 5 } 12

cos 538 5 } c 12 cos 538

12(tan 538) 5 b

c5}

15.92 ø b

C

a

B

32

a

tan 648 5 } 32

cos 648 5 } c 32 cos 648

32(tan 648) 5 a 188

a

c

B 5 908 2 648 5 268

c ø 19.94

24

648

C

23. A b

cos 498 5 } 27

27(sin 498) 5 a 538


27

b

B 5 908 2 498 5 418

b

c5}

65.61 ø a

B

c ø 73.0

28. A

A 5 908 2 188 5 728 a

cos 188 5 } 24

b

b

24(sin 188) 5 b

C

sin 188 5 } 24

24(cos 188) 5 a 22.83 ø a

7.42 ø b

10.8 a

248

B

A 5 908 2 248 5 668 a

24. A 678

C

c ø 57.96

26. A

sin 358 5 } 16

7

c5}

b

cos 248 5 } 10.8 c a

a tan 678 5 }7

7(tan 678) 5 a 16.49 ø a

10.8(cos 248) 5 a

10.8(sin 248) 5 b

9.87 ø a

4.39 ø b

B

29. a.

B 5 908 2 678 5 238 7 cos 678 5 }c 7 cos 678

c5} c ø 17.92

sin 248 5 } 10.8

opp sinu 5 } hyp x

sin 308 5 }1 1 2

}5x

P 5 6(2x) 5 12x 5 12 1 }2 2 5 6 1

The perimeter of the hexagon is 6 units.


n2ws13-a.indd 717

717

6/27/06 11:39:43 AM

Chapter 13,

continued

b. n-sided polygon

water’s surface

33. 250 ft

360 8 Interior angle: 1 } n 2

1500 ft 158

180 8 Half of interior angle: 1 } n 2

h

(180n (8 (180n (8 one triangle out of n-sided polygon

1

1500(tan 158) 5 h 401.92 ø h The depth of the submersible after the dive is about 250 1 402 5 652 feet.

x

1

1808 Length of half of side: x 5 sin } n

2

As the angle of the dive increases, the final depth increases.

Perimeter: P 5 n p 2x 180 8 P 5 2n p sin1 } n 2

34. a. 6315 ft

c. As n gets larger, the perimeter of the polygon

approaches the circumference of the circle. Because r 5 1, C 5 2π.

a 2 11,200 ft

29.48

a 2 11,200

b.

180 8 P 5 2n p sin1 } n 2

sin 29.48 5 } 6315 6315 sin 29.48 5 a 2 11,200

180 8 2π 5 2n p sin1 } n 2

6315 sin 29.48 1 11,200 5 a 14,300 ø a

180 8 π 5 n sin1 } n 2

The altitude of the ridge is about 14,300 feet.

1

2

180 8 When n 5 50: π 5 50 sin } ø 3.14 50

35. a.

Tropic of Cancer r 23.58

Problem Solving 30.

R

23.58 R

h

tan 318 5 } 25

h

25(tan 318) 5 h

Equator

318 25 ft

15.02 ø h

R 5 3960

The height of the tree is about 15 feet. 31.

r

cos 23.58 5 } R

x sin 258 5 } 150

r

cos 23.58 5 } 3960

150(sin 258) 5 x

3960(cos 23.58) 5 r

63.39 ø x The length of the prop holding open the piano is about 63.4 cm.

The circumference of the Tropic of Cancer is about 22,818 miles.

32. h

b. Diameter of Tropic of Cancer:

368 138 ft

D 5 2r 5 2(3631.56) 5 7263 h

tan 368 5 } 138

138 l

cos 368 5 } 138 cos 368

138(tan 368) 5 h

l5}

100.26 ø h

l ø 170.58

The railway’s height is about 100 feet and its length is about 171 feet.

718

3631.56 ø r C 5 2πr 5 2π(3631.56) ø 22,818

The distance between two points is about 7263 miles.


1

h

tan 158 5 } 1500


n2ws13-a.indd 718

6/27/06 11:39:51 AM

Chapter 13, 36. a.

continued tan 328 tan 538

50 tan 328 1 h } 5 h

158 258

u 25,000 ft

328 1 tan tan 538 2

d

50 tan 328 5 h 2 h }

x

y

1

25,000 cos 658 5 } d

50 tan 328

25,000 cos 658

59 ø h

The distance from the airplane to the first town is about 59,155 feet. x

3 yr 365 days 24 h 60 min 60 sec 38. } p } p } p } p } 1 1 yr 1 day 1h 1 min

25,000(tan 658) 5 x 53,613 ø x

5 94,608,000 seconds

The horizontal distance from the airplane to the first town is about 53,613 feet. x1y

tan 758 5 } 25,000 25,000(tan 758) 5 x 1 y 93,301 ø x 1 y 93,301 ø 53,613 1 y 39,688 ø y The distance between the towns is about 39,688 feet. This was found by calculating the horizontal distance, x 1 y, between the airplane and the second town and subtracting the length of x, 53,613 feet, to find y, the distance between the two towns.


The building is about 59 meters high. Mixed Review

tan 658 5 } 25,000

c.

2

}5h tan 328 12} tan 538

d 5 } ø 59,155

b.

tan 328 tan 538

50 tan 328 5 h 1 2 }

u 5 908 2 258 5 658

37.

10 pt 1 qt 1 gal 5 1 39. } p } p } 5 }, or 1} gallons 4 1 2 pt 4 qt 4 500 500 ft 1 yd 2 40. } p } 5 }, or 166 } yards 3 1 3 3 ft 9.4 kg 1000 g 41. } p } 5 9400 grams 1 1 kg 2 tons 2000 lb 16 oz 42. } p } p } 5 64,000 ounces 1 1 ton 1 lb 43. 5.6 meters to millimeters

5.6 m 1

1000 mm 1m

} p } 5 5600 millimeters

44.

2x 2 y 5 8 → y 5 28 1 2x 3x 1 4y 5 23 3x 1 4(28 1 2x) 5 23 3x 2 32 1 8x 5 23

h 328 50 m

11x 2 32 5 23

538 x

11x 5 55 x55

h

tan 538 5 }x

y 5 28 1 2(5) 5 2 The solution is (5, 2).

h x5} tan 538

45. 4x 1 3y 5 21

h

tan 328 5 } 50 1 x h

4x 1 3(3 2 3x) 5 21

tan 328 5 } h

4x 1 9 2 9x 5 21

tan 538

25x 1 9 5 21

50 1 }

1

3x 1 y 5 3 → y 5 3 2 3x

h tan 538

2

tan 328 50 1 } 5 h

25x 5 210 x52 y 5 3 2 3(2) 5 23 The solution is (2, 23). 46. 9x 2 12y 5 20

32

18x 2 24y 5 40

26x 1 8y 5 11

33

218x 1 24y 5 33 0 5 77 No solution


n2ws13-a.indd 719

719

6/27/06 11:39:58 AM

Chapter 13, 47. 12x 1 5y 5 4

continued 24x 1 10y 5 8

32

3x 2 10y 5 28

52. y 2 2 4y 1 16x 1 116 5 0

A 5 0, B 5 0, C 5 1

3x 2 10y 5 28

B 2 2 4AC 5 0 2 2 4(0)(1) 5 0

50

27x

Because B 2 2 4AC 5 0, the conic is a parabola.

x50

y 2 2 4y 1 16x 1 116 5 0

3(0) 2 10y 5 28

( y 2 2 4y 1 4) 5 216x 2 116 1 4

210y 5 28

( y 2 2)2 5 216x 2 112

4

y 5 }5

1

( y 2 2)2 5 216(x 1 7)

2

4 The solution is 0, }5 . 48. 22x 1 7y 5 16

53. 9x 2 1 25y 2 1 162x 1 250y 1 454 5 0

3x 2 5y 5 22

A 5 9, B 5 0, C 5 25

26x 1 21y 5 48

33

B 2 2 4AC 5 0 2 2 4(9)(25) 5 2900

6x 2 10y 5 24

32

Because B 2 2 4AC < 0 and A Þ C, the conic is an ellipse.

11y 5 44 y54

9x 2 1 25y 2 1 162x 1 250y 1 454 5 0 9(x 2 1 18x 1 81) 1 25( y 1 10y 1 25)

3x 2 5(4) 5 22 3x 5 18

5 2454 1 729 1 625

x56

9(x 1 9)2 1 25( y 1 5)2 5 900 (x 1 9)2 100

The solution is (6, 4). 49. 4x 2 6y 5 18

22x 1 y 5 23 → y 5 23 1 2x

54. x 2 1 8x 1 4y 1 28 5 0

A 5 1, B 5 0, C 5 0

4x 2 6(23 1 2x) 5 18

B 2 2 4AC 5 0 2 2 4(1)(0) 5 0

4x 1 18 2 12x 5 18

Because B 2 2 4AC 5 0, the conic is a parabola.

28x 5 0

x 2 1 8x 1 4y 1 28 5 0

x50

(x 2 1 8x 1 16) 5 24y 2 28 1 16

y 5 23 1 2(0) 5 23

(x 1 4)2 5 24( y 1 3)

50. x 2 1 y 2 1 4x 1 6y 2 17 5 0

55. x 2 2 y 2 1 14x 1 16y 2 5 5 0

A 5 1, B 5 0, C 5 1

A 5 1, B 5 0, C 5 21

B 2 2 4AC 5 0 2 2 4(1)(1) 5 24

B 2 2 4AC 5 0 2 2 4(1)(21) 5 4

2

Because B 2 4AC < 0, B 5 0, and A 5 C, the conic is a circle.

Because B 2 2 4AC > 0, the conic is a hyperbola.

2

x 2 2 y 2 1 14x 1 16y 2 5 5 0

x 1 y 1 4x 1 6y 2 17 5 0

(x

2

1 4x 1 4) 1 ( y 1 6y 1 9) 5 17 1 4 1 9

(x 2 1 14x 1 49) 2 ( y 2 2 16y 1 64) 5 5 1 49 2 64

2

2

2

(x 1 7)2 2 ( y 2 8)2 5 210

(x 1 2) 1 ( y 1 3) 5 30 2

( y 2 8)2 10

2

51. x 2 4y 1 6x 1 16y 1 137 5 0

A 5 1, B 5 0, C 5 24 B 2 2 4AC 5 0 2 2 4(1)(24) 5 16 Because B 2 2 4AC > 0, the conic is a hyperbola. x 2 2 4y 2 1 6x 1 16y 1 137 5 0

(x

2

1 6x 1 9) 2 4( y 2 4y 1 4) 5 2137 1 9 2 16 2

(x 1 3)2 2 4( y 2 2)2 5 2144 (x 1 3)2 ( y 2 2)2 }2}51 144 36

(x 1 7)2 10

}2}51

Lesson 13.2 13.2 Guided Practice (pp. 860–862) 1.

y

658 x


(x 1 4)2 5 24y 2 12

The solution is (0, 23).

2

( y 1 5)2 36

}1}51

Sample answer: 658 1 3608 5 4258 658 2 3608 5 22958 A positive coterminal angle is 4258. A negative coterminal angle is 22958.

720


n2ws13-a.indd 720

6/27/06 11:40:05 AM

Chapter 13, 2.

continued 2. If the angle is positive, you travel counter-clockwise from

y

the initial position on the x-axis. If the angle is negative, you travel clockwise from the initial position.

2308 x

3. B;

Because 22408 is negative, the terminal side is 2408 clockwise from the positive x-axis. Sample answer:

4. C;

Because 6008 is 2408 more than 3608, the terminal side makes one whole revolution counterclockwise plus 2408 more.

2308 1 3608 5 5908 2308 2 3608 5 21308 A positive coterminal angle is 5908. A negative coterminal angle is 21308. 3.

5. A;

9π

1808 1 π radians 2

} 5 24058 2} 4

y

Because 24058 is negative, the terminal side makes one whole revolution clockwise plus 458 more.

x

3008

6.

7.

y

y

1108

Sample answer:

x x

3008 1 3608 5 6608

2108

3008 2 3608 5 2608 A positive coterminal angle is 6608. A negative coterminal angle is 2608. 4.

8.

9.

y

y

y

4508 x

x

x

29008

7408

1


2

1808 10. (6π radians) } 5 10808 π radians

Sample answer: 7408 2 2(3608) 5 208

y

7408 2 3(3608) 5 23408 A positive coterminal angle is 208. A negative coterminal angle is 23408.

x

135π 3π πradians 5. 1358 5 1358 } 5 } 5 } radians 180 4 1808

1

2

5π πradians 6. 2508 5 2508 } 5 2} radians 18 1808

1

2

21

6π

21

1

2

5π 5π 1808 7. } 5 } radians } 5 2258 4 4 π radians

1

y 5π 18

π 1808 π 8. } 5 1 } radians 2 } 5 188 10 10 π radians

1

2

π 9. r 5 220, u 5 } 2

π

Arc length: s 5 ru 5 2201 } 5 110π ø 346 22 1

1

π

Area: A 5 }2 r 2u 5 }2 (220)21 } 5 12,100π ø 38,013 22 The length of the outfield fence is about 346 feet, and the area of the field is about 38,013 square feet.

2

5π 1808 11. } radians } 5 508 18 π radians

x

21

2

25π 1808 12. } radians } 5 23008 3 π radians

1

y 25π 3

x

13.2 Exercises (pp. 862–865) Skill Practice 1. An angle is in standard position if its vertex is at the

origin and its initial side lies on the positive x-axis.


n2ws13-a.indd 721

721

6/27/06 11:40:16 AM

Chapter 13, 21

1

continued 1808 π π 27. } 5 1 } radians 2 } 5 208 9 9 π radians

2

1

26π 1808 13. } radians } 5 5208 9 π radians y

π π 1808 28. 2} 5 1 2} radians 2 } 5 2458 4 4 π radians

1

1

14. C; The terminal side makes one and a half revolutions

counterclockwise plus a few more degrees. 5708 5 3608 1 1808 1 308

1

21

2

1 13π 6

1808 2 1 π radians 2

14π 1808 14π 30. } 5 } radians } 5 1688 15 15 π radians 31. B;

13π 6

} 5 } radians } 5 3908

15. Sample answer:

π 32. r 5 4 in., u 5 } 6

708 1 3608 5 4308 708 2 3608 5 22908

π

2π

s 5 ru 5 41 } 5} ø 2.09 3 62

16. Sample answer:

2558 1 3608 5 6158

The arc length is about 2.09 inches.

2558 2 3608 5 21058

1

π

1

4π

A 5 }2 r 2u 5 }2 (4)21 } 5} ø 4.19 3 62

17. Sample answer:

21258 1 3608 5 2358

The area of the sector is about 4.19 square inches.

21258 2 3608 5 24858

5π 33. r 5 3 m, u 5 } 12 5π 5π s 5 ru 5 3 } 5} ø 3.93 4 12

18 Sample answer:

1 2

8208 2 2(3608) 5 8208 2 7208 5 1008 8208 2 3(3608) 5 8208 2 10808 5 22608

The arc length is about 3.93 meters.

19. Sample answer:

1

1

15π

5π

A 5 }2 r 2u 5 }2 (3)21 } 5} ø 5.89 8 12 2

9π 9π 8π π 9π } 2 2(2π) 5 } 2 4π 5 } 2 } 5 } 2 2 2 2 2

The area of the sector is about 5.89 square meters.

9π 9π 3π 12π 9π } 2 3(2π) 5 } 2 6π 5 } 2 } 5 2} 2 2 2 2 2

34. r 5 15 cm, u 5 458

1 π radians 2

45π

π

458 } 5 } 5} radians 180 4 1808

20. Sample answer:

7π

7π

12π

7π

7π

12π

π

5π

s 5 ru 5 15 1 } ø 11.78 42

2} 1 2π 5 2} 1} 5} 6 6 6 6

The arc length is about 11.78 centimeters.

19π

2} 2 2π 5 2} 2} 5 2} 6 6 6 6

1

28π 9

18π 9

28π 9

28π 9

36π 9

20π 3

20π 3

20π 3

18π 3

20π 3

20π 3

20π 3

24π 3

8π 9

22. Sample answer:

2π 3

} 2 3(2π) 5 } 2 6π 5 } 2 } 5 }

4π 3

} 2 4(2π) 5 } 2 8π 5 } 2 } 5 2}

2π π radians 23. 408 5 408 } 5 } radians 9 1808

2


2

13π π radians 25. 22608 5 22608 } 5 2} radians 9 1808

1

2


1

2

2π

35. r 5 12 ft, u 5 1508

} 2 2(2π) 5 } 2 4π 5 } 2 } 5 2}

1

π

The area of the sector is about 88.36 square centimeters.

10π 9

} 2 2π 5 } 2 } 5 }

1

1

A 5 }2 r 2u 5 }2 (15)21 } 5} ø 88.36 8 42

21. Sample answer:

28π 9

2

1808 29. 5π 5 (5π radians) } 5 9008 π radians

x

28π 9

2

1 π radians 2 150π 5π 5π s 5 ru 5 12 1 } 5 10π ø 31.42 6 2

1508 } 5 } 5} 180 6 1808

The arc length is about 31.42 feet. 1

1

5π

A 5 }2 r 2u 5 }2 (12)21 } 5 60π ø 188.5 6 2 The area of the sector is about 188.5 square feet. 36. r 5 18m, u 5 258

1 π radians 2 5π 5π 5π s 5 ru 5 18 1 } 5} ø 7.85 2 36 2

258 } 5 } radians 36 1808


26π 9

2

The arc length is about 7.85 meters. 1

1

5π

45π

A 5 }2 r 2u 5 }2 (18)21 } 5} ø 70.69 2 36 2 The area of the sector is about 70.69 square meters.

722


n2ws13-a.indd 722

6/27/06 11:40:23 AM

Chapter 13,

continued

37. r 5 25 in., u 5 2708

50.

3π π radians 2708 } 5 } radians 2 1808

1

11

2

10

1

before substituting it into the formula.

2

1 23608 1h 2

Angle of minute hand: 8 h } 5 228808 }

Ï2 π 40. sin } 5 sin 458 5 } 2 4

}

1 π 43. cot 1 } 2 5 } π ø 2.41 8 tan 1 } 82 3π 45. sin } ø 0.975 7

1 2

1 π 42. sec } 5 } π ø 1.06 9 cos 1 } 92 } Ï3 π 44. cos } 5 cos 308 5 } 2 6 1 4π 46. csc } 5 } ø 1.35 4π 15 sin 1 } 2 15

u 47. v 5 }t 3608

3608

u

3608

1 π radians 1808 2

34.928 5 34.928 } 5 0.194π radians


s 5 ru 5 21.16(0.194π) ø 12.9

60 min 1h

p} Second hand: v 5 }t 5 } 1 min

The arc would be about 12.9 meters long.

5 21,6008 per hour

1 1 b. A 5 } r 2u 5 } (21.16)2(0.194π) ø 136.44 2 2

The area of the sector is about 136.44 square meters.

Problem Solving

π 53. a. s 5 ru 5 42 1 } 2 ø 16.49 8

48. In one day, Earth rotates 3608.

3608 24 h

4 h p } 5 608

The length of the arc formed by the outer edge of a step is about 16.49 inches.

π In 4 hours, Earth rotates 608 or } radians. 3 3608 π radians 49. 200 rev } 5 72,0008 5 72,0008 } 1 rev 1808

2

15 rev 2π rad 1 min π radians 15 rev 51. a. } 5 } p } p } 5 } 2 sec 1 min 1 rev 60 sec 1 min π } The people rotated 2 radians each second. 58 π b. d 5 58ft, u 5 }, r 5 } 5 29 ft 2 2 π s 5 ru 5 29 1 } ø 45.6 22

52. a. r 5 21.16m, u 5 34.928

5 3608 per hour Minute hand: v 5 }t 5 } 1h

1

The minute hand travels 12 times farther than the hour hand.

The people rotated through an arc length of about 45.6 feet each second.

5 308 per hour Hour hand: v 5 }t 5 } 12 h u

4π

4π

2π

Ï3 π 41. tan } 5 tan 308 5 } 3 6

1 π radians 1808 2

radians. or 2} 3

A 5 }2 r 2u 5 }2 (6)21 } 5 4π ø 12.57 cm 2 9 2 1 π 39. cos } 5 cos 608 5 } 2 3

u 5 2908 1 (21508) 5 22408

The angle generated by the hour hand is 22408,

40π 2π π radians 408 } 5 } 5} radians 180 9 1808

u

5

5 22408 } 5 2} radians 3

38. The angle measure 408 was not converted to radians

1

6

1875π

3π

The area of the sector is about 1472.62 square inches.

1

4 7

A 5 }2 r 2u 5 }2 (25)2 } 5} ø 1472.62 4 2

1

3 8

The arc length is about 117.81 inches. 1

1 22408 2

9

75π

s 5 ru 5 251 } 5} ø 117.81 2 2 2 3π

12

1

2

5 400π radians A point on the CD spins through an angle of 72,0008, or 400π radians, in one minute.

π b. Because each step has a central angle of } and there 8

are 15 steps plus a landing, the staircase will cover π

161 } or 2π, which is equivalent to 3608. 82

1 1 π c. area of one step: A 5 } r 2u 5 } (42)2 1 } 2 2 2 8 441π

ø 346.36 in.2 5} 4 Area of 15 steps ø 15(346.36) ø 5195.4 in.2 You would need about 5195.4 square inches of carpeting to cover the 15 steps.


n2ws13-a.indd 723

723

6/27/06 11:40:29 AM

Chapter 13,

continued

54. a. 20 sectors on dartboard

63.

π radians 1808

y

π

u5} 5 1881 } 2 5 } radians 20 10 3608

(7, 2)

2

1 2

Area of entire sector: A 5 } r 2u

x

2

5 2 π

1

A 5 }2 1 6}8 2 1 } 10 2 2809π

A5} ø 6.89 1280

220

7

3 2 π

A 5 6.89 2 }2 1 6 }8 2 }8 2 1 } 10 2 5

Equation of tangent line: y 2 2 5 2}2 (x 2 7)

2π

1

7

1

5 2}2 Slope of tangent line at (7, 2) 5 2} m

Area of double region: 1

2

Slope of radius: m 5 } 5 }7 720

The area of the entire sector is about 6.89 square inches.

1 A 5 6.89 2 }2 1 6 }4 2 } 10

7

49

7

53

y 2 2 5 2}2 x 1 } 2

625π

y 5 2}2 x 1 } 2

A 5 6.89 2 } ø 0.75 320 The area of the double region is about 0.75 square inches.

64.

y

(2, 6)

Area of triple region:

2

π

π 3 2 1 3 1 3 2} A 5 }2 1 3 }4 1 }8 2 1 } 3} 10 2 2 } 2 1 4 2 1 10 2

x

2

π

π 1 1 2 1 3 2} A 5 }2 1 4 }8 2 1 } 3} 10 2 2 } 2 1 4 2 1 10 2 225π

620

Slope of radius: m 5 } 53 220

The area of the triple region is about 0.46 square inches.

1

Equation of tangent line: y 2 6 5 2}3 (x 2 2)

The probability that it lands in the double region is about 0.11.

1

2

1

20

1

5

y 2 6 5 2}3 x 1 }3

Area of triple 0.46 in.2 P(triple) 5 }} 5 }2 ø 0.07 Area of sector 6.89 in.

The probability that it lands in the triple region is about 0.07.

1

1

5 2}3 Slope of tangent line at (2, 6) 5 2} m

Area of double 0.75 in.2 b. P(double) 5 }} 5 }2 ø 0.11 Area of sector 6.89 in.

y 5 2}3 x 1 } 3 65.

y

(5, 11)

Mixed Review }

}

}

}

55. Ï 54 5 Ï 9 p Ï 6 5 3Ï 6 }

}

}

3

3

}

56. Ï 320 5 Ï 64 p Ï 5 5 8Ï 5 }

}

}

}

}

x

}

57. Ï 36 p Ï 18 5 6 p Ï 9 p Ï 2 5 6 p 3Ï 2 5 18Ï 2 }

}

}

}

}

}

58. Ï 3 p Ï 60 5 Ï 180 5 Ï 36 p Ï 5 5 6Ï 5 }

}

}

}

}

}

11 2 0

Ï5 Ï5 5 }5} } 5 }

}

5 2} Slope of tangent line at (5, 11) 5 2} m 11

}

}

}

}

}

}

}

}

}

}

}

Ï84 Ï 12 Ï 7 Ï 4 p Ï 21 2Ï21 Ï 12 61. } } p } } 5 } } 5 } 5 } 5 } 7 7 7 Ï7 Ï7 Ï7 }

}

}

}

}

Ï4 p Ï7 Ï 14 Ï7 Ï7 Ï2 Ï28 62. } } } 5 } } 5 } } p } } 5 } } 5 } 2 Ï4 p Ï2 Ï2 Ï2 Ï2 Ï8

724

11

Slope of radius: m 5 } 5} 5 520

Ï49 Ï49 7 Ï27 Ï9 Ï3 3Ï3 27 60. } 5 } 5 } 5 } Ï64 Ï64 8 8 59.

5

(x 2 5) Equation of tangent line: y 2 11 5 2} 11 5


1089π

A5} 2} ø 0.46 1280 320

25

y 2 11 5 2} x1} 11 11 5

146

y 5 2} x1} 11 11


n2ws13-a.indd 724

6/27/06 11:40:38 AM

Chapter 13,

continued

8! 8! 8 p 7 p 6 p 5! 66. 8P3 5 } 5 } 5 } 5 336 5! 5! (8 2 3)!

4. B

7! 7! 7 p 6 p 5 p 4 p 3! 67. 7P4 5 } 5 } 5 }} 5 840 3! 3! (7 2 4)!

a

12! 12! 12 p 11 p 10! 68. 12P2 5 } 5 } 5 } 5 132 10! 10! (12 2 2)!

C

10! 10! 10 p 9 p 8 p 7 p 6 p 5 p 4 p 3 p 2! 69. 10P8 5 } 5 } 5 }}} 2! 2! (10 2 8)!

428

b

15! 15! 71. 15C7 5 } 5 } 5 6435 8!7! (15 2 7)!7!

b

13.38 ø a

c 158 b

A

B 5 908 2 158 5 758 9

9

tan 158 5 }b

sin 158 5 }c

9 tan 158

9 sin 158

b 5 } ø 33.59

B

a

508 b A

C

c 5 } ø 34.77

6. B

c

14

B 5 908 2 508 5 408 14 tan 508 5 } b 14 } b5 ø 11.75 tan 508

378

C

14 sin 508 5 } c 14 } c5 ø 18.28 sin 508

12

b

A

A 5 908 2 378 5 538 a

b

cos 378 5 } 12

sin 378 5 } 12

12(cos 378) 5 a

2. B


18(cos 428) 5 a

12.04 ø b

C

Quiz 13.1–13.2 (p. 865)

a

c

C

258 10

A

B 5 908 2 258 5 658 10

a

tan 258 5 } 10

cos 258 5 } c 10 cos 258

c 5 } ø 11.03

10(tan 258) 5 a 4.66 ø a

12sin 378 5 b

9.58 ø a

7.22 ø b

7. Sample answer:

8. Sample answer:

1158 1 3608 5 4758

2908 1 3608 5 6508

1158 2 3608 5 22458

2908 2 3608 5 2708

9. Sample answer:

10. Sample answer:

22π 4π } 1 2π 5 } 9 9

7π 5

14π 9

4π 9

708

c b

7π 5

3π 5

} 2 2π 5 2}

11. r 5 8 in., u 5 1158

1 π radians 2

A

5

tan 208 5 }b b 5 } ø 13.74

23π

1158 5 1158 } 5 } radians 36 1808

A 5 908 2 708 5 208

5 tan 208

17π 5

} 1 2π 5 }

} 2 2π 5 2}

3. B

C

cos 428 5 } 18

18(sin 428) 5 b

9

8! 8! 73. 8C4 5 } 5 } 5 70 4!4! (8 2 4)!4!

5

a

sin 428 5 } 18

5. B

6! 6! 72. 6C5 5 } 5 } 5 6 1!5! (6 2 5)!5!

1.

A

A 5 908 2 428 5 488

5 1,814,400 9! 9! 70. 9C5 5 } 5 } 5 126 4!5! (9 2 5)!5!

18

5

sin 208 5 }c 5 sin 208

c 5 } ø 14.62

46π

s 5 ru 5 81 } 5} ø 16.06 9 36 2 23π

1

1

184π

A 5 }2 r 2u 5 }2 (8)2 1 } 5} ø 64.23 9 36 2 23π

The arc length of the sector is about 16.06 inches, and the area is about 64.23 square inches.


n2ws13-a.indd 725

725

6/27/06 11:40:47 AM

Chapter 13,

continued

12.

4. u 5 1808 152 ft

y

h

308

u

h sin 308 5 } 152

(21, 0)

r

x

152(sin 308) 5 h 152 1 }2 2 5 h 1

x 5 21, y 5 0 }

}

r 5 Ïx 2 1 y 2 5 Ï(21)2 1 0 2 5 Ï1 5 1

76 5 h The height of the escalator is 76 feet.

y

0

x

21

y

0

}

sin u 5 } 5 }1 5 0 r

Lesson 13.3

5 21 cos u 5 }r 5 } 1


50 tan u 5 }x 5 } 21

1. x 5 3, y 5 23 }

}

r 5 Ïx 2 1 y 2 5 Ï3 2 1 (23)2 5 Ï18 5 3Ï2 }

}

1

r

csc u 5 }y 5 }0 undefined

}

y Ï2 23 1 sin u 5 } 5} } 5 2} } 5 2} r 2 Ï2 3Ï2

r

1

x

21

5 21 sec u 5 }x 5 } 21

}

Ï2 3 x 1 cos u 5 }r 5 } } 5 } } 5 } 2 Ï2 3Ï2

undefined cot u 5 }y 5 } 0

y 23 5 21 tan u 5 }x 5 } 3

5.

y

}

} 3Ï2 r 5 2Ï2 csc u 5 }y 5 } 23 }

r

3Ï2

x

3

2108 x

u9

}

5 Ï2 sec u 5 }x 5 } 3 5 21 cot u 5 }y 5 } 23 2. x 5 28, y 5 15 }

}}

r 5 Ïx 2 1 y 2 5 Ï(28)2 1 15 2 5 Ï289 5 17 y

}

6.

y

u9

15

sin u 5 } 5} r 17

x

22608

28 8 x 5 2} cos u 5 }r 5 } 17 17 y

15

r

17

r

17

u is coterminal with 1008, whose terminal side lies in Quadrant II. So, u 9 5 1808 2 1008 5 808.

15

5 2} tan u 5 }x 5 } 28 8 7.

csc u 5 }y 5 } 15

y

17

5 2} sec u 5 }x 5 } 28 8

x

u9

27π 9

28 8 x 5 2} cot u 5 }y 5 } 15 15 3. x 5 25, y 5 212 }

11π

}}

u is coterminal with } , whose terminal side lies in 9

r 5 Ïx 2 1 y 2 5 Ï(25)2 1 (212)2 5 Ï169 5 13

y 212 12 sin u 5 } 5} 5 2} r 13 13 25 5 x } } } cos u 5 r 5 13 5 213 y 212 12 tan u 5 }x 5 } 5} 25 5 13 13 r csc u 5 }y 5 } 5 2} 212 12 13 13 r sec u 5 }x 5 } 5 2} 25 5 25 5 x } } } cot u 5 y 5 212 5 12

726

}

11π

2π

Quadrant III. So, u 9 5 } 2π5} . 9 9 8.

y

u9

x


The terminal side of a lies in Quadrant III. So, u 9 5 2108 2 1808 5 308.

15π 4

7π

u is coterminal with } , whose terminal side lies in 4 7π

π

Quadrant IV. So, u 9 5 2π 2 } 5} . 4 4


n2ws13-a.indd 726

6/27/06 11:40:58 AM

Chapter 13, 9.

continued 4. x 5 29, y 5 12

y

}

u9

22108

12

x

29

3

y

12

4

r

15

r

15

5

x

29

3

4

5 2}5 cos u 5 }r 5 } 15

u is coterminal with 1508, whose terminal side lies in Quadrant II. So, u 9 5 1808 2 1508 5 308. In Quadrant II, cosine is negative.

5 2}3 tan u 5 }x 5 } 29 5

5 }4 csc u 5 }y 5 } 12

}

Ï3 2

So, cos (22108) 5 2cos 308 5 2}.

5 2}3 sec u 5 }x 5 } 29

27 ft.

10. u 5 208, v 5 } sec.

5 2}4 cot u 5 }y 5 } 12

Model for horizontal distance: v2

5. x 5 27, y 5 224

d5} sin 2u 32

}

}}

r 5 Ïx 2 1 y 2 5 Ï(27)2 1 (224)2 5 Ï625 5 25

(27)2 d5} sin(2 p 208) 32

y

224

x

27

y

224

r

25

r

25

25

x

27

7

}

24

sin u 5 } 5} 5 2} r 25 25

729

d5} sin(408) ø 14.64 32 The track and field long jumper can jump a horizontal distance of about 14.64 feet. y

sin u 5 }r

7

5 2} cos u 5 }r 5 } 25 25 24

5} tan u 5 }x 5 } 27 7 25

5 2} csc u 5 }y 5 } 224 24

y

sin 1008 5 } 5.25

5 2} sec u 5 }x 5 } 27 7

5.25(sin 1008) 5 y 5.17 ø y Copyright © by McDougal Littell, a division of Houghton Mifflin Company.

y

}

sin u 5 } 5} 5 }5 r 15

x

11.

}}

r 5 Ïx 2 1 y 2 5 Ï(29)2 1 12 2 5 Ï225 5 15

5} cot u 5 }y 5 } 224 24

The top of the treadmill is about 6 1 5.17 5 11.17 feet about the ground.

6. x 5 5, y 5 212 }

}}

13.3 Exercises (pp. 870–872)

r 5 Ïx 2 1 y 2 5 Ï5 2 1 (212)2 5 Ï169 5 13

Skill Practice

sin u 5 } 5} 5 2} r 13 13

1. A reference angle is an angle in standard position whose

terminal side lies on an axis. 2. If u is in Quadrant III, subtract 1808 from u to determine

the reference angle. Then, compute the cosine of this angle and make it negative. 3. x 5 8, y 5 15 }

}

y

15

x

8

y

15

cos u 5 }r 5 } 17 tan u 5 }x 5 } 8

212

x

5

y

212

12

r

13

13

12

cos u 5 }r 5 } 13 5 2} tan u 5 }x 5 } 5 5 5 2} csc u 5 }y 5 } 212 12 r

13

x

5

sec u 5 }x 5 } 5

r 5 Ïx 2 1 y 2 5 Ï8 2 1 15 2 5 Ï289 5 17 sin u 5 } 5} r 17

y

}

}

r

17

r

17

x

8

csc u 5 }y 5 } 15

5

5 2} cot u 5 }y 5 } 212 12

sec u 5 }x 5 } 8 cot u 5 }y 5 } 15


n2ws13-a.indd 727

727

6/27/06 11:41:04 AM

Chapter 13,

continued }

10. x 5 5, y 5 2Ï 11

7. x 5 2, y 5 22 }

}

}

}

}

y

Ï2 22 1 Ï2 2Ï2 } Ï2 x 2 1 cos u 5 }r 5 } } 5 } } 5 } 2 Ï2 2Ï 2 y 22 5 21 tan u 5 }x 5 } 2

sin u 5 } 5} } 5 2} } 5 2} r 2

}

r

2Ï2

r

2Ï2

x

2

}

}

}

2

}

x

26 2 Ï13 3Ï13 y 9 3 5 2}2 tan u 5 }x 5 } 26

x

26

Ï 11

r

6 2Ï11

6Ï11

r

6

x

5 2Ï11

}

}

}

}

5Ï11

y

4

24

tan a 5 }x 5 } 5 }7 27 }

2Ï13

12.

cos u 5 }r 5 } } 5 2} } 5 2} 13

3Ï13

2Ï11

x 5 27, y 5 24

}

y 9 3 3Ï13 sin u 5 } 5} } 5 } } 5 } r 13 Ï13 3Ï13

r

y

11. C;

r 5 Ïx 1 y 5 Ï(26) 1 9 5 Ï117 5 3Ï13

}

5

cot u 5 }y 5 } } 5 2} 11 2

3Ï13 9

x

Ï 11

sec u 5 }x 5 }5

}

r y

2Ï11

csc u 5 }y 5 } } 5 2} 11

8. x 5 26, y 5 9 2

y

5 2} tan u 5 }x 5 } 5 5

5 21 cot u 5 }y 5 } 22 2

}

}

cos u 5 }r 5 }6

5 Ï2 sec u 5 }x 5 } 2

}

}

sin u 5 } 5} 5 2} r 6 6

5 2Ï2 csc u 5 }y 5 } 22 }

}} } 2

}

r 5 Ïx 2 1 y 2 5 Ï5 2 1 (2Ï11 ) 5 Ï36 5 6

r 5 Ïx 2 1 y 2 5 Ï2 2 1 (22)2 5 Ï8 5 2Ï2

y

u 5 08 (1, 0) x

}

Ï13 3

csc u 5 } 5 } 5 } }

Ï13

x 5 1, y 5 0, r 5 1

2

9. x 5 23, y 5 25 }

y

0

x

1

y

0

sin u 5 } 5 }1 5 0 r

5 2}3 cot u 5 }y 5 } 9 }}

cos u 5 }r 5 }1 5 1

r 5 Ïx 2 1 y 2 5 Ï(23)2 1 (25)2 5 Ï34

}

tan u 5 }x 5 }1 5 0

}

y 25 5Ï34 sin u 5 } 5} } 5 2} r 34 Ï34 }

23 3Ï34 x cos u 5 }r 5 } } 5 2} 34 Ï34 y

25

r

Ï34

Ï34

r

Ï34

}

Ï34

x

23

13.

y

(0, 1)

r

1

x

1

r

1

r

1

x

0

sec u 5 }x 5 }1 5 1 cot u 5 }y 5 }0 undefined

π

5

x }

5 2} csc u 5 }y 5 } 25 5 }

5 2} sec u 5 }x 5 } 23 3 3

5 }5 cot u 5 }y 5 } 25

x 5 0, y 5 1, r 5 1 y

1

x

0

y

1

sin u 5 } 5 }1 5 1 r cos u 5 }r 5 }1 5 0 tan u 5 }x 5 }0 undefined

728

1

u52

5 }3 tan u 5 }x 5 } 23 }

r

csc u 5 }y 5 }0 undefined

csc u 5 }y 5 }1 5 1 sec u 5 }x 5 }0 undefined cot u 5 }y 5 }1 5 0


}

5 2} sec u 5 }x 5 } 26 2


n2ws13-a.indd 728

6/27/06 11:41:14 AM

Chapter 13, 14.

y

continued 19.

u 5 5408

y

23708 (21, 0)

x x

u9

x 5 21, y 5 0, r 5 1 y 0 sin u 5 } 5 }1 5 0 r

r 1 csc u 5 }y 5 }0 undefined

x 21 5 21 cos u 5 }r 5 } 1

r 1 sec u 5 }x 5 } 5 21 21

y

0

50 tan u 5 }x 5 } 21 15.

y

u5

x

u is coterminal with 3508, whose terminal side lies in Quadrant IV. So, u 9 5 3608 2 3508 5 108. 20.

21

y

u9

cot u 5 }y 5 } undefined 0

x 25π 6

7π 2

7π

u is coterminal with } , whose terminal side lies in 6 Quadrant III. So, u 9 5 } 2π5} . 6 6

(0, 21)

21.

x 5 0, y 5 21, r 5 1 y 21 sin u 5 } 5} 5 21 r 1 x

0

y

21

cos u 5 }r 5 }1 5 0 undefined tan u 5 }x 5 } 0 16.

r

y 8π 3

u9

1

csc u 5 }y 5 } 5 21 21 r

1

x

0

x

sec u 5 }x 5 }0 undefined 2π


cot u 5 }y 5 }1 5 0

π

2π

Quadrant II. So, u 9 5 π 2 } 5} . 3 3

y

22.


π

7π

x

y

x

u9 21008

u9 15π 4

u is coterminal with 2608, whose terminal side lies in Quadrant III. So, u 9 5 2608 2 1808 5 808. 17.

x

7π


y

π

7π

u9

Quadrant IV. So, u 9 5 2π 2 } 5} . 4 4

1508 x

23.

y 213π 6

The terminal side of u lies in Quadrant II. So, u 9 5 1808 2 1508 5 308. 18.

y

u9 x

11π

u is coterminal with } , whose terminal side lies in 6 11π

3208

u9

x

π

Quadrant IV. So, u 9 5 2π 2 } 5} . 6 6

The terminal side of u lies in Quadrant IV. So, u 9 5 3608 2 3208 5 408.


n2ws13-a.indd 729

729

6/27/06 11:41:28 AM

Chapter 13, 24.

continued 29.

y

y

u 5 1358

u52

u9 x

x

u9

π

4π

4π

u 9 5 1808 2 1358 5 458.

u is coterminal with } . So, u 9 5 } 2π5} . 3 3 3

Secant is negative in Quadrant II. } So, sec 1358 5 2sec 458 5 2Ï2 . 25.

8π 3

Cotangent is positive in Quadrant III. }

π

Ï3

So, cot1 2} 5 cot 1 } 5} . 3 2 3 32 8π

y

u 5 2408

30.

y

x

u9

u9

u52

3π 4

x

u 9 5 2408 2 1808 5 608. Tangent is positive in Quadrant III. } So, tan 2408 5 tan 608 5 Ï3 . 26.

5π

5π

4π

π

u is coterminal with } . So, u 9 5 } 2} 5} . 4 4 4 4 Tangent is positive in Quadrant III.

y

π

So, tan1 2} 5 tan } 5 1. 4 2 4 3π

x

u9

31.

y

u 5 21508

u9

u5

Sine is negative in Quadrant III. 1

So, sin(21508) 5 2sin 308 5 2}2 . 27.

11π

π

u 9 5 2π 2 } 5} . 6 6

y

Secant is positive in Quadrant IV.

u 5 24208 u9

π

11π

}

2Ï 3

So, sec } 5 sec } 5} . 3 6 6

x

y x 32. The equation for tangent is tanu 5 }, not tanu 5 }y . x 3 So, tanu 5 }4 .

u is coterminal with 3008. So, u 9 5 3608 2 3008 5 608. Cosecant is negative in Quadrant IV.

}

2Ï3

1 2 1

opp

}

}

hyp

}

opp hyp hyp sinu 33. tanu 5 } } 5} 5} adj 1 cosu adj }

So, csc(24208) 5 2csc 608 5 2} . 3 28.

x

11π 6

hyp

sin 908 1 tan 908 5 } 5 }0 undefined cos 908

y

cos 908 sin 908

0

cot 908 5 } 5 }1 5 0 u5

7π 4

u9

x

34.

e a 1 bi 5 e a(cos b 1 i sin b) ln e a 1 bi 5 ln[e a(cos b 1 i sin b)]

7π

(a 1 bi) ln e 5 ln e a 1 ln(cos b 1 i sin b)

π

u 9 5 2π 2 } 5} . 4 4

a 1 bi 5 a ln e 1 ln(cos b 1 i sin b)

Cosine is positive in Quadrant IV. }

Ï2 7π π So, cos } 5 cos } 5} . 2 4 4

a 1 bi 5 a 1 ln(cos b 1 i sin b)


u is coterminal with 2108. So, u 9 5 2108 2 1808 5 308.

bi 5 ln(cos b 1 i sin b) e bi 5 e ln(cos b 1 i sin b) e bi 5 cos b 1 i sin b

730


n2ws13-a.indd 730

6/27/06 11:41:40 AM

Chapter 13,

continued

When b 5 π:

37.

y

πi

e 5 cos π 1 i sin π e πi 5 21 1 i(0) e πi 5 21 e πi 1 1 5 0 Problem Solving

Stop

75 ft

y

v2 d5} sin 2u 32 (49)2 5} sin(2 p 458) 32 2401

5} sin 908 32 2401

5} (1) ø 75.03 ft 32 Distance of friend’s kick: v2

d5} sin 2u 32 (49)2 5} sin(2 p 608) 32 2401

5} sin 1208 32

75 ft Start 10 ft

Ground

u is coterminal with 1658. So, u 9 5 1808 2 1658 5 158. y

sin 158 5 } 75 75 sin 158 5 y 19.4 ø y When the ride stops, you are about 10 1 75 1 19.4 5 104.4 feet above the ground. If the radius is doubled, your height above the ground is not doubled because your starting height is not doubled. 38. a. u 5 1178

u 9 5 1808 2 1178 5 638 y

sin 638 5 } 128 128 sin 638 5 y

}

1 2

2401 Ï3 } ø 64.98 ft 5} 32 2

Difference ø 75.03 2 64.98 5 10.05


Your football will travel about 10 feet farther than your friend’s football. 36.

114 ø y

}}}

}}

5 Ï47,629.21 ø 218.2

160 5 v 2 sin 368 160 } 5 v2 sin 368

272.21 ø v 2 16.5 ø v The in-line skater’s speed should be about 16.5 feet per second.

258.1 ø x

d 5 Ï(258.1 2 128)2 1 (114 2 0)2

v2

v2 sin 368 55} 32

128 cos 638 5 2x

The coordinates are about (258.1, 114).

5 Ï(2186.1)2 1 (114)2

55} sin(2 p 188) 32

x

cos 638 5 2} 128

b. (128, 0), (258.1, 114)

v2

d5} sin 2u 32

x

u 9 5 158

35. v 5 49 feet per second

Distance of your kick:

u 5 1658

}

The distance between the centers of the two unbonded oxygen atoms is about 218.2 picometers. 39. a. v 5 25 ft. per sec.

v2

d5} sin 2u 32

(25)2

u 5 258: d 5 } sin(2 p 258) ø 15 32 (25)2

u 5 308: d 5 } sin(2 p 308) ø 16.9 32 (25)2

u 5 358: d 5 } sin(2 p 358) ø 18.4 32 (25)2

u 5 408: d 5 } sin(2 p 408) ø 19.2 32 (25)2

u 5 458: d 5 } sin(2 p 458) ø 19.5 32 (25)2

u 5 508: d 5 } sin(2 p 508) ø 19.2 32 (25)2

u 5 558: d 5 } sin(2 p 558) ø 18.4 32 (25)2

u 5 608: d 5 } sin(2 p 608) ø 16.9 32


n2ws13-a.indd 731

731

6/27/06 11:41:44 AM

Chapter 13,

continued

(25)2

Mixed Review

u 5 658: d 5 } sin(2 p 658) ø 15 32 u

258

d

15.0 16.9 18.4 19.2 19.5 19.2 18.4

u

608

d

16.9 15.0

308

358

408

458

508

41. f(x) 5 5x 1 2

558

f(x) 2

658

b. A value of u 5 458 appears to maximize the horizontal v2 v2 } is constant sin 2 u , distance. In the formula d 5 } 32 32

in this situation. So, d is maximized when sin 2u is as large as possible. The maximum value of sine occurs at 908, so, 2u 5 908 and u 5 458. (0, 1)

x

21

Because no horizontal line intersects the graph of f more than once, the inverse of f is a function. 42. f(x) 5 2x 1 7 f(x)

y

908 x

1

x

21

c. When u 5 (45 2 k)8:

y

2

v d5} sin[2 p (45 2 k)8] 32

Because no horizontal line intersects the graph of f more than once, the inverse of f is a function.

(90 1 2k)8 (90 2 2k)8

43. f(x) 5 x 2 1 5

v2 5} sin(90 2 2k)8 32

u9

f(x)

u9 x

When u 5 (45 1 k)8: v2

d5} sin[2 p (45 1 k)8] 32 v2

5} sin(90 1 2k)8 32

40. Circle of

x

C

latitude

x

21

Because a horizontal line intersects the graph of f more than once, the inverse of f is not a function. 44. f(x) 5 4x 2, x ≥ 0 f(x)

P

R

u O

1

R

E

Equator

1 2 1 x 5 }2 R

2πx 5 } (2πR)

x

21

Because x ≥ 0, no horizontal line intersects the graph of f more than once. The inverse of f is a function. 45. f(x) 5 0.25x 2 f(x)

1

}R x 2 1 5} 5 }2 cosu 5 } R R

u 5 608 The circumference of the circle of latitude at P is half the distance around the equator at latitude 608.

1 21


1

The reference angles for (90 1 2k)8 and (90 2 2k)8 will always be equal and will be in Quadrants I and II or in Quadrants III and IV. Because sine is positive in Quadrants I and II and negative in Quadrants III and IV, the values of sin(90 1 2k)8 and sin(90 2 2k)8 will always be equal. So, the distances are the same.

x

Because a horizontal line intersects the graph of f more than once, the inverse of f is not a function.

732


n2ws13-a.indd 732

6/27/06 11:41:56 AM

Chapter 13,

continued 18

46. f(x) 5 x 2 7

52.

f(x)

18

18

∑ (4i 1 1) 5 4i∑ i1 ∑1 i51 51 i51

1

18(18 1 1)

2

54 } 1 18 2 5 4(171) 1 18 5 702 24

53.

1

5 408 2 2(300) 5 2192

Range 5 11 2 2 5 9

Ï

1 2 rn

Sn 5 a11 } 12r2

8

(2 2 5.375)2 1 . . . 1 (11 2 5.375)2

}

Ï

65.875

1 1 2 (3) 2 5

s 5 }}} 5 } 8 8

S5 5 2 } 5 242 123

ø 2.87

7

48. 4, 7, 9, 12, 13, 15, 17, 18

55.

Range 5 18 2 4 5 14 8

8

Ï

1 2 rn

1 2 1 2 1 2 1 2

3 7 1 2 }2 1 } } S7 5 4 3 1 2 }2

168.875 } ø 4.59 8

2187 } 1 1 2 128 } } 54 1 2}2


49. 1.8, 2.9, 3.7, 5.7, 5.9, 6.1, 8.2

Range 5 8.2 2 1.8 5 6.4 1.8 1 2.9 1 3.7 1 5.7 1 5.9 1 6.1 1 8.2 34.3 }x 5 }}}} 5 } 5 4.9 7

7

22059

}

1 128 5 }4 } 1 2}2

}}}}

s5

Ï

(1.8 2 4.9)2 1 (2.9 2 4.9)2 1 . . . 1 (8.2 2 4.9)2 }}}} 7

2059

}

Ï

`

50. 51, 53, 54, 56 57, 59, 60, 62

56.

Range 5 62 2 51 5 11 8

a1

5 56.5 }}}}

(51 2 56.5) 1 (53 2 56.5) 1 . . . 1 (62 2 56.5) }}}} 8 2

i21

a1 5 8, r 5 }2

8

2

∑ 8 1 }1 2 i51 2

1

51 1 53 1 54 1 56 1 57 1 59 1 60 1 62 452 }x 5 }}}} 5}

Ï

11

5} 5 8} 256 256

29.02

5 } ø 2.04 7

s5

3

Sn 5 a11 } 12r2

(4 2 11.875) 1 (7 2 11.875) 1 . . . 1 (18 2 11.875)2 }}}}} 8 2

}

5

i21

1

}}}}}

2

∑ }1 1 }3 2 i514 2

a1 5 }4, r 5 }2, n 5 7

4 1 7 1 9 1 12 1 13 1 15 1 17 1 18 95 }x 5 }}} 5 } 5 11.875

Ï

∑ 2(3)i 2 1

i51

a1 5 2, r 5 3, n 5 5

2 1 3 1 3 1 4 1 5 1 7 1 8 1 11 43 }x 5 }}} 5 } 5 5.375 }}}

2

5

54.

8

24(24 1 1)

5 17(24) 2 2 } 2

Because a horizontal line intersects the graph of f more than once, the inverse of f is not a function. 47. 2, 3, 3, 4, 5, 7, 8, 11

24

1

x

21

s5

24

∑ (17 2 2i) 5 17i∑ 122∑i i51 51 i51

2

8

S5} 5 }1 5 16 12r 1 2 }2

}

Ï

98

5 } 5 3.5 8 15

51.

15

15

∑ (3i 1 2) 5 3i∑ i1 ∑2 i51 51 i51

1

15(15 1 1)

2

15

53 } 12∑1 2 i51

5 3(120) 1 2(15) 5 390


n2ws13-a.indd 733

733

6/27/06 11:42:01 AM

Chapter 13,

continued

Mixed Review of Problem Solving (p. 873)

100 ft

u 5 3008 d

x u9

negative in Quadrants II and IV. So, an angle for which the secant is positive and the cotangent is negative lies in Quadrant IV.

Goal line

y

Sample answer: 3008

(50, 0)

5. a. From 7:00 P.M. to 8:55 P.M. the restaurant completed 55 (3608) 5 3308. one revolution plus } 60

x

50 ft

(x, y)

π radians 1808

u 5 6908 5 6908 1 } 2 5 } radians 6

u 9 5 3608 2 3008 5 608

s 5 ru

x cosu 9 5 }r x cos 608 5 } 50

s 5 47.25 1 } ø 569 6 2 23π

You revolve about 569 feet. b. No; The radius of their circle is only 42.25 feet.

50(cos 608) 5 x 50 1 }2 2 5 x 1

6. a.

Distance to goal 5 d 1 100 5 25 1 100 5 125 You are 125 feet from the goal line when you leave the circle. π radians 5π b. u 5 3008 5 3008 } 5 } radians 3 1808

1

2

778

100 b. cos 778 5 } d

100 m

100 cos 778

The straight-line distance between you and your friend is about 444.54 meters.

5π

s 5 50 1 } 3 2

π π radians 7. a. u 5 608 5 608 } 5 } radians 3 1808

1

250

s5} π ø 262 3

1

You march around the circle about 262 feet. h sin 708 5 } 400

Kite

π radians 1808

1

The kite is about 376 1 4 5 380 feet above ground.

400 ft

You

x

400(cos 708) 5 x 137 ø x

π

1

π

u 5 458 5 458 1 } 2 5 } radians 4

376 ø h

cos 708 5 } 400

2

A 5 }2 r 2u 5 }2 (6)2 1 } 5 6π ø 18.85 32

400(sin 708) 5 h

858

708 x

1

π

49π

A 5 }2 r 2u 5 }2 (7)2 1 } 5} ø 19.24 8 42

h

Friend y

376

tan 858 5 } y 376 tan 858

y5} y ø 33

You and your friend are standing about 137 1 33 5 170 feet apart.

The area of the first slice is about 18.85 square inches and the area of the second slice is about 19.24 square inches. $1.50 b. First slice: }2 5 $.08 per square inch 18.85 in. $1.70

5 $.09 per square inch Second slice: } 19.24 The first slice is the better buy. c. Charge $.08 per square inch.

$.08 in.

2 Price 5 } 2 p 19.24 in. 5 $1.54

Decrease the price of the 7-inch slice to $1.54.

y

u 5 5608 u9

The height of the Sears Tower is about 433.15 meters.

h

d 5 } ø 444.54

s 5 ru

3.

d

433.15 ø h

d 5 50 2 x 5 50 2 25 5 25

b.

h

tan 778 5 } 100 100(tan 778) 5 h

25 5 x

2. a.

23π


1. a.

4. Secant is positive in Quadrants I and IV. Cotangent is

x

u is coterminal with 2008, whose terminal side lies in Quadrant III. So, u 9 5 2008 2 1808 5 208.

734


n2ws13-a.indd 734

6/27/06 11:42:09 AM

Chapter 13,

continued

Lesson 13.4

5. cos u 5 0.4

u 5 cos210.4 ø 66.48

Investigating Algebra Activity 13.4 (p. 874) 1.

u

2π

π

3π

π

2} 2} 4 2

2} 4

}

f (u) 5 sinu

0

Ï2 2}

Ï2 2}

g (u ) 5 cosu

21

2} 2

0

Ï2 }

3π 4

π

2

2

}

}

0

Ï2 }

1

Ï2 }

}

66.48

2

1

Ï2 }

2

2

In Quadrant IV, u ø 3608 2 66.48 5 293.68. 6. tan u 5 2.1

u 5 tan212.1 ø 64.58

g (u ) 5 cosu

0

2} 2

y

0

2

}

Ï2

x

}

}

f (u ) 5 sinu

66.48

u

}

}

Ï2

π 2

u

}

}

21

y

π 4

0

64.58

u

21

x

64.58

2. f(u ) 5 sin u does not have an inverse over the domain

2π ≤ u ≤ π, because more than one value of u maps to } π

Ï2

3π

and } map to } . the same y-value. For example, both } 4 4 2 3. g(u ) 5 cos u does not have an inverse over the domain

2π ≤ u ≤ π, because more than one value of u maps to π

In Quadrant III, u ø 1808 1 64.58 5 244.58. 7. sin u 5 20.23

u 5 sin21(20.23) ø 213.38

π

y

the same y-value. For example, both 2} and } map to 0. 2 2 4. The graphs fail the horizontal line test.

u

x

Draw Conclusions

213.38

1. Sample answer:

π

π Domain: 2} ≤u≤} 2


2

In this domain, sin u does not have any repeated values. 2. Sample answer:

In Quadrant IV, u ø 3608 2 13.38 5 346.78. 8. tan u 5 4.7

u 5 tan214.7 ø 788 y

Domain: 0 ≤ u ≤ π In this domain, cos u does not have any repeated values. 3. Sample answer:

No. Because f(u ) 5 sin u and g(u ) 5 cos u are periodic functions, there are an infinite number of domains in which the functions have inverses. 13.4 Guided Practice (pp. 876–877) π π 1. When 2} ≤ u ≤ }, or 2908 ≤ u ≤ 908: 2 2 }

Ï2

In Quadrant III, u ø 1808 1 788 5 2588. 9. sin u 5 0.62

u 5 sin210.62 ø 38.38 y

π

u5

π 5} or 608 3 2

x

788

u 5 sin21 } 5} or 458 4 2 2. When 0 ≤ u ≤ π, or 0 ≤ u ≤ 1808:

788

u

38.38

u

38.38 x

1 cos21 }

π π 3. When 2} < u < }, or 2908 < u < 908: 2 2 π u 5 tan21(21) 5 2} or 2458 4

In Quadrant II, u ø 1808 2 38.38 5 141.78.

π π 4. When 2} ≤ u ≤ }, or 2908 ≤ u ≤ 908: 2 2 π 1 u 5 sin21 2}2 5 2} or 2308 6

1 2


n2ws13-a.indd 735

735

6/27/06 11:42:18 AM

Chapter 13,

continued 10. When 0 ≤ u ≤ π or 08 ≤ u ≤ 1808:

10. cos u 5 20.39

u 5 cos21(20.39) ø 1138 y

1 Ï2

u

u 5 cos21 } } 5 458

x

678

1

11. C; When 08 ≤ u ≤ 1808:

1138

678

2π

u 5 cos211 2}2 2 5 } or 1208 3

12. u 5 sin21 0.18 ø 0.18 ø 10.48 13. u 5 tan21 2.6 ø 1.20 ø 69.08

In Quadrant III, u ø 1808 1 678 5 2478. 4 11. cos u 5 } 9

u5

4 cos21 }

ø 63.68

9

10 5 12. tan u 5 } 5 } 8 4 5 21 } u 5 tan 4 ø 51.38

5 13. sin u 5 } 12

15. u 5 cos21(20.4) ø 1.98 ø 113.68 16. u 5 tan21(20.75) ø 20.64 ø 236.98 17. u 5 sin21(20.2) ø 20.20 ø 211.58 18. u 5 sin21 0.8 ø 0.93 ø 53.18

5

u 5 sin21 } ø 24.68 12 14.

14. u 5 cos21 0.36 ø 1.20 ø 68.98

19. u 5 cos21 0.99 ø 0.14 ø 8.18 20. u 5 cos21(20.82) ø 145.18 y

26 ft

10 ft

145.18 u

u

10

34.98 34.98

5

sin u 5 } 5} 26 13

x

5 u 5 sin21 } ø 22.68 13

1 2

The angle of the ramp is about 22.68. 13.4 Exercises (pp. 878 – 880)

In Quadrant III, u ø 1808 1 34.98 ø 214.98. 21. u 5 sin21(20.45) 5 226.78 y

Skill Practice

2. tan21 3 is defined because the range of tan u is all real

numbers. The range of cos u is 21 ≤ cos u ≤ 1, so cos21 3 is undefined because 3 is not in the range.

π π 3. When 2} ≤ u ≤ } or 2908 ≤ u ≤ 908: 2 2 π u 5 sin21 1 5 } or 908 2 π π 4. When 2} ≤ u ≤ } or 2908 ≤ u ≤ 908: 2 2 π 21 u 5 tan (21) 5 2} or 2458 4

u 26.78

26.78 x

In Quadrant III, u ø 1808 1 26.78 ø 206.78. 22. u 5 sin210.15 5 8.68 y

u 8.68

8.68

x

5. When 0 ≤ u ≤ π or 08 ≤ u ≤ 1808:

π

u 5 cos21 0 5 } or 908 2 6. There is no angle whose cosine is 22. So, cos21(22)

In Quadrant II, u ø 1808 2 8.68 ø 171.48. 23. u 5 tan21 3.2 ø 72.68 y

is undefined. π π 7. When 2} ≤ u ≤ } or 2908 ≤ u ≤ 908: 2 2 }

1 2

π

Ï3 u 5 sin21 } 5} or 608 3 2

π π 8. When 2} ≤ u ≤ } or 2908 ≤ u ≤ 908: 2 2 π 1 21 } u 5 sin 2 5 } or 308 6 π π 9. When 2} ≤ u ≤ } or 2908 ≤ u ≤ 908: 2 2 π 1 u 5 tan21 2} } 5 2} or 2308 6 Ï3

1

736

u 72.68

72.68 x

In Quadrant III, u ø 1808 1 72.68 ø 252.68.


1 π 1. The inverse sine of } is }, or 308 2 6

2


n2ws13-a.indd 736

6/27/06 11:42:28 AM

Chapter 13,

continued

24. u 5 tan21(25.3) ø 279.38

33. Let tan21 x 5 y

x 5 tan y

y

cot(tan

21

u

79.38

34.

x

79.38

1

1

21 } x) 5 cot y 5 } tan y 5 x 5 x

Let cos21 x 5 y x 5 cos y 1

In Quadrant II, u ø 1808 2 79.38 ø 100.78. 25. u 5 cos21 0.25 ø 75.58

1

sec(cos 21 x) 5 sec y 5 } 5 x21 cos y 5 } x Problem Solving 35.

y

75.58

u

100 ft

x

75.58

80 ft

90 ft

u 10 ft

In Quadrant IV, u ø 3608 2 75.58 ø 284.58.

80

in the domain.

u 5 sin211 }5 2 ø 53.18 4

y

u

44.48

4

sin u 5 } 5 }5 100

26. The domain restriction is 908 < u < 1808 and 44.48 is not

The firefighter should set the ladder at an angle of about 53.18.

44.48

36.

x

angle of descent

x

u

31,000 ft

In Quadrant II, u ø 1808 2 44.48 ø 135.68. 5 27. sin u 5 } 8

6 28. tan u 5 } 1


5

u 5 sin211 }8 2 ø 38.78

104 miles 5 549,120 ft

Not drawn to scale

u 5 tan21 6 ø 80.58

5280 feet 104 miles p } 5 549,120 ft 1 mile

29. 4

549,120

tan x 5 } 31,000

u

x

runway

2 3

1

549,120

2

x 5 tan21 } ø 86.778 31,000 }

2Ï3

}

Ï3

u ø 908 2 86.778 5 3.238

cos x 5 } 5} 4 2 }

1 Ï3 2

The airplane descends at an angle of about 3.238.

x 5 cos21 } 5 308 2

37.

11 ft

u 5 908 1 x 5 908 1 308 5 1208

u 17 ft

30. Cosine is positive in Quadrants I and IV.

Sine is negative in Quadrants III and IV. If cos u > 0 and sin u < 0, then u is located in Quadrant IV. Sample answer: u 5 2458 31. Sine is negative in Quadrants III and IV.

Tangent is positive in Quadrants I and III. If sin u < 0 and tan u > 0, then u is located in Quadrant III. Because 3608 ≤ u ≤ 720, u must be between 5408 and 6308. Sample answer: u 5 6008 32.

Let sin21 x 5 y x 5 sin y

11

tanu 5 } 17

u 5 tan211 } ø 338 17 2 11

The angle of repose is about 338. Because the angle of repose remains the same, you can use u 5 338 to find the radius of the 15 foot high pile. 15 ft

1

1

csc(sin21 x) 5 csc y 5 } 5 }x 5 x 21 sin y

338 r


n2ws13-a.indd 737

737

6/27/06 11:42:39 AM

Chapter 13,

continued

15 tan 338

15

tan 338 5 } → r 5 } ø 23 r

40.

The diameter is about d 5 2r 5 2(23) 5 46 feet. 38.

3 ft

u1 4 ft

u2

x

a.

A

u

x

y

d

B

4

1

A 5 tan211 }x 2; B 5 tan211 }x 2

sin u1

4

1

u 5 A 2 B 5 tan21}x 2 tan21}x

} 5 1.333

sin u2

sin 708 sin u2

r

1 ft

u3

1.333 1

}5}

sin 708

ø 0.705 sin u2 5 } 1.333 Maximum X=1.9999987 Y=36.869898

u2 5 sin210.705 ø 44.88 x b. tan 44.88 5 } 4

On a graphing calculator, the maximum occurs at x ø 2 and u ø 378. So, the camera should be positioned about 2 feet from the wall.

x 5 4 tan 44.88 ø 3.97 Because vertical angles are congruent:

Mixed Review

u1 5 u2 1 u3

41.

708 5 u2 1 u3 y

8x 1 32 5 22x 2 3

tan 708 5 }4

10x 5 235

y 5 4 tan 708 ø 11 The distance x is about 3.97 feet and the distance y is about 11 feet. d 5 y 2 x ø 11 2 3.97 5 7.03

35

1

x 5 2} 5 23 }2 10 42. 18x 2 1 x 2 5 5 0

9x 1 5 5 0

d. As you get closer to the goggles, the distance d

will get smaller. u1 and u2 will also decrease until you are standing over the goggles. At this point, d will equal zero.

or

2x 2 1 5 0

9x 5 25

or

2x 5 1

5 x 5 2}9

or

x 5 }2

12x 2 5 8x 1 15

43.

d

1

2

12x 2 8x 2 15 5 0

100 ft

(6x 1 5)(2x 2 3) 5 0

u

30 mi

5280 ft 1 mi

1h 60 min

1 min

6x 1 5 5 0

44 ft

v5} p}p}p} 5} 1 sec 1h 60 sec d 5 vt 5 44t d

44t

tan u 5 } 5} 100 100

2x 2 3 5 0

6x 5 25

or

2x 5 3

5 x 5 2}6

or

3

x 2 2 10x 2 3 5 0

44t

}}

1 100 2 to

44t tan21 }

2(210) 6 Ï(210)2 2 4(1)(23) 2(1)

x 5 }}} }

10 6 Ï112

}

5} 5 5 6 2Ï 7 2

738

1

x 5 }2 5 1}2

44. 3x 2 2 30x 2 9 5 0

u 5 tan211 } 100 2 You need to turn your head an angle of u 5 see the cyclist t seconds later.

or


(9x 1 5)(2x 2 1) 5 0

c. The distance d is about 7.03 feet.

39.

3

1

x 1 4 5 2}4 x 2 }8


n2ws13-a.indd 738

6/27/06 11:42:51 AM

Chapter 13,

continued 10 2 51. } 5 } x x26

45. 27x 3 2 64 5 0

27x 3 5 64

2x 5 10(x 2 6)

64

x3 5 } 27

Î3 } x3 5

2x 5 10x 2 60

Î6427

28x 5 260

}

3

}

4

60

1

x 5 }3 5 1}3 46.

1Î x 1 12

23

2

10

}26

}

2

10

4 3

4 3

0} Check: } 15 15

Î3 } x 1 12 5 5 3 }

15

x5} 5} 8 2

2

3

5 (5)

}0} 3 15 } } 2 2

x 1 12 5 125 x 5 113 47.

}5}✓

(6x 2 11)5/2 5 243

1 (6x 2 11)5/2 22/5 5 (243)2/5 6x 2 11 5 9

29 3 52. } 5 } 4x 21x

3(4x) 5 29(2 1 x)

6x 5 20 20 6

12x 5 218 2 9x

1 3

x 5 } 5 3} 48.

8

x24

2

21x 5 218 18

3x 2 8

5 32

6

x 5 2} 5 2}7 21

23(x 2 4) 5 25(3x 2 8)

3

29

Check: } 0 }

1 2

3(x 2 4) 5 5(3x 2 8) 3x 2 12 5 15x 2 40 1

x5} 5 2}3 212


49.

102x 2 6 5 12 102x 5 18

3

29

21 8

21 8

}5}✓

53.

5 x11

27 x11

}135}

log 102x 5 log 18

5 1 3(x 1 1) 5 27

2x(log 10) 5 log 18

5 1 3x 1 3 5 27 3x 5 215

2x(1) 5 log 18 log 18 x5} ø 0.628 2 50.

6

}0} 8 224 } } 7 7

212x 5 228 228

41 2}7 2

6 7

2 1 2}

4 x

x 5 25 5 27 Check: } 130} 25 1 1 25 1 1

9 x15

}5}

5

}5}✓

20 5 5x 54.

45x 9

Check: }4 0 } 415 4 4

7 4

7 4

4x 1 20 5 9x

4

7

2}4 1 3 0 }4

4(x 1 5) 5 9x

9 9

}0}

151✓

1 x13

x 3x 1 16

}5}

3x 1 16 5 x(x 1 3) 3x 1 16 5 x 2 1 3x 0 5 x 2 2 16 0 5 (x 1 4)(x 2 4) 24 5 x

or

45x

Check: x 5 24 1 24 1 3

24 3(24) 1 16

}0}

21 5 21 ✓


n2ws13-a.indd 739

739

6/27/06 11:42:56 AM

Chapter 13,

continued 3.

Check: x 5 4:

3x 5 2x 2 8 1 12 3(4) 12 Check: } 021} 424 424 12 0

8 4Ï5

2 Ï5

2Ï 5

x

4 4Ï 5

1 Ï5

Ï5

y

8

r

4Ï 5

r

4Ï 5

x

4

}

}

}

Ï5

}

}

1

cot u 5 }y 5 }8 5 }2 4.

Undefined

x 5 212, y 5 23 }

}}

}

x 5 6, y 5 22 }

}

}

}

1 Ï17

x

212 3Ï 17

4 Ï17

y

23

r

3Ï 17

r

3Ï 17

x

212

Ï17

}

4Ï17

1

5 }4 tan u 5 }x 5 } 212

}

y Ï10 22 1 sin u 5 }r 5 } } 5 2} } 5 2} 10 Ï10 2Ï10

}

}

5 2Ï17 csc u 5 }y 5 } 23

}

6 3 3Ï 10 x cos u 5 }r 5 } } 5 } } 5 } 10 Ï10 2Ï10

}

}

Ï17

5 2} sec u 5 }x 5 } 212 4

1

5 2}3 tan u 5 }x 5 } 6

54 cot u 5 }y 5 } 23

}

csc u 5 } 5 } 5 2Ï 10 }

23 3Ï17

cos u 5 }r 5 } } 5 2} } 5 2 } 17

r 5 Ïx 2 1 y 2 5 Ï 62 1 (22)2 5 Ï 40 5 2Ï10

}

y

}

sin u 5 }r 5 } } 5 2} } 5 2} 17

Quiz 13.3–13.4 (p. 880)

5.

}

Ï10 2Ï10 r 5} sec u 5 }x 5 } 6 3 6 x cot u 5 }y 5 } 5 23 22

y

u 5 1508 u9 x

x 5 27, y 5 5 }

}

}

r 5 Ïx 2 1 y 2 5 Ï(27)2 1 52 5 Ï 74 y r

5 Ï74

u 9 5 1808 2 1508 5 308

}

5Ï 74 74

sin u 5 } 5 } } 5 }

Cosine is negative in Quadrant II. }

Ï3

}

27 7Ï 74 x cos u 5 }r 5 } } 5 2} 74 Ï74 y

5

5

5 2}7 tan u 5 }x 5 } 27 r

. So, cos 1508 5 2cos 308 5 2} 2 6.

y

u5

u9

}

Ï74

8π 3

csc u 5 }y 5 } 5

x

}

r

Ï74

x

27

}

r 5 Ï x2 1 y2 5 Ï(212)2 1 (23)2 5 Ï153 5 3Ï17

There is no solution.

}

Ï74

52} sec u 5 }x 5 } 27 7 7

5 2}5 cot u 5 }y 5 } 5

2π

u is coterminal with } , whose terminal side lies in 3 Quadrant II. 2π

π

So, u 9 5 π 2 } 5} . 3 3 Tangent is negative in Quadrant II. 8π

2π


12 0

2.

y

5 Ï5 sec u 5 }x 5 } 4

}021}

2Ï10 22

}

5} csc u 5 }y 5 } 8 2

x54

r y

}

}

tan u 5 }x 5 }4 5 2

3x 5 2(x 2 4) 1 12

22

}

cos u 5 }r 5 } } 5 } } 5 } 5

12 3x 55. } 5 2 1 } x24 x24

y

}

sin u 5 }r 5 } } 5 } } 5 } 5

1 1 }5}✓ 7 7

1.

x 5 4, y 5 8

r 5 Ïx2 1 y2 5 Ï42 1 82 5 Ï80 5 4Ï5

1 4 }0} 413 3(4) 1 16

}

5 2tan } 5 2Ï 3 . So, tan } 3 3

740


n2ws13-b.indd 740

6/28/06 2:09:36 PM

Chapter 13, 7.

continued 14. u 5 tan21 6 ø 80.58

y

y

x

u9

80.58

u

u 5 28408

x

80.58

u is coterminal with 2408, whose terminal side lies in Quadrant III. So, u 9 5 2408 2 1808 5 608. Sine is negative in Quadrant III. }

In Quadrant III, u ø 1808 1 80.5 ø 260.58. 15. u 5 cos21(20.72) ø 136.18

Ï3

8.

y

. So, sin(28408) 5 2sin 608 5 2} 2 u52

15π 4

y

136.18

u9

x

x

u is coterminal with 458, whose terminal side lies in π

Quadrant I. So, u 9 5 } . 4

u ø 136.18 16. u 5 sin21(20.55) ø 233.48 y

Secant is positive in Quadrant I. π

5 sec } 5 Ï2 . So, sec 1 2} 4 2 4 15π

u

}

33.48 x

9. When 0 ≤ u ≤ π or 08 ≤ u ≤ 1808: }

1

Ï2

2

3π

u 5 cos21 2} 5} or 1358 2 4 π π 10. When 2 } ≤ u ≤ } or 2908 ≤ u ≤ 908: 2 2

In Quadrant IV, u ø 3608 2 33.48 ø 326.68. 17.

L

30 ft

π


u 5 sin21(21) 5 2} or 2908 2 π π 11. When 2} < u < } or 2908 < u < 908: 2 2

1368 d 2

30 ft

h

d

}

u5

π 5} or 308 6 3

Ï3 tan21 }

12. When 0 ≤ u ≤ π or 08 ≤ u ≤ 1808:

1

π

u 5 cos21 }2 5 } or 608 3 13. u 5 sin21 0.30 ø 17.58 y

Ground

L

cos 688 5 } 30 L 5 30 cos 688 ø 11.2 h ø 30 2 11.2 ø 18.8 The stuntman was about 18.8 feet above the ground at the beginning and end of the swing. d 2 } }

d

sin 688 5 30 5 } 60

u 17.58

30 ft

688

17.58

x

d 5 60 sin 688 ø 55.6 The two platforms are about 55.6 feet apart.

In Quadrant II, u ø 1808 2 17.58 ø 162.58.

Lesson 13.5 Investigating Algebra Activity 13.5 (p. 881) 1. Answers will vary. The values of the ratios will be

the same. 2. Answers will vary. The ratios are always equal for

any triangle.


n2ws13-b.indd 741

741

6/28/06 2:10:13 PM

Chapter 13,

continued


18 sin 1228

c sin 23.68

}5}

B

1.

18 sin 23.68 sin 1228

c 5 } ø 8.5 348

c

B ø 34.48,

a

C ø 23.68,

c ø 8.5

4. Because A is acute: 1008

h 5 b sin A 5 12 sin 368 ø 7.1

C

A 5 1808 2 348 2 1008 5 468 8 a }5} sin 468 sin 348

Because 7.1 < 9 < 12 (h < a < b), there are two triangles. Triangle 1: C 12

a(sin 348) 5 8 sin 468 8 sin 468 sin 348

A

a 5 } ø 10.3 }5}

B ø sin 210.7837 ø 51.68

8 sin 1008 sin 348

a ø 10.3,

c ø 14.1

B

A

518

C ø 1808 2 51.682368 ø 92.48 c 9 }5} sin 92.48 sin 368 9 sin 92.48 sin 368

c5}

448

11

B

12 sin 368

c 5 } ø 14.1 2.

c

ø 0.7837 sin B 5 } 9

c(sin 348) 5 8 sin 1008

A 5 468,

368

12 9 }5} sin B sin 368

8 sin 348

c sin 1008

a

a

b

c ø 15.3 B ø 51.68,

C

C ø 92.48,

Triangle 2:

C 5 1808 2 518 2 448 5 858

C

11 a }5} sin 858 sin 518

12

a(sin 858) 5 11 sin 518 11 sin 518 sin 858 11 b }5} sin 858 sin 448

A

a 5 } ø 8.6

368 c B

B ø 1808 2 51.68 ø 128.48 C ø 1808 2 368 2 128.48 ø 15.6

11 sin 448 sin 858

b 5 } ø 7.7 a ø 8.6,

9

In Triangle 1, B ø 51.68. The obtuse angle B has 51.68 as a reference angle.

b(sin 858) 5 11 sin 448

C 5 858,

b ø 7.7

3. Because A is obtuse and a > b, there is exactly one

triangle.

c sin 15.68

12 sin 128.48

}5}

12 sin 15.68 sin 128.48

c5} c ø 4.1

C

B ø 128.48,

A

1228

C ø 15.68,

c ø 4.1

5. Because a is acute:

18

12

18 sin 1228

c ø 15.3

h 5 b sin A 5 4 sin 508 ø 3.06 Because 3.06 > 2.8 (h > a), there is no triangle. c

B

12 sin B

}5}


8

A

12 sin 1228

sin B 5 } ø 0.5654 18 B ø 34.48 C ø 1808 2 1228 2 34.48 ø 23.6

742


n2ws13-b.indd 742

6/27/06 11:50:04 AM

Chapter 13,

continued

6. Because B is obtuse and b > a, there is one triangle.

5. A 5 528,

C

b 5 42

Because A is acute:

13

6 B

1058

c

h 5 b sin A 5 42 sin 528 ø 33.1

A

Because 33.1 > 32 (h > a), there is no triangle.

13 sin 1058

6 sin A

6. A 5 378,

}5}

a 5 8,

b 5 14

This is the SSA case.

65 sin 1058

ø 0.4458 sin A 5 } 13

Because A is acute:

A ø sin 0.4458 ø 26.58

h 5 b sin A 5 14 sin 378 ø 8.4

C ø 1808 2 1058 2 26.58 ø 48.58

Because 8.4 > 8 (h > a), there is no triangle.

21

7. A 5 288,

13 c }5} sin 1058 sin 48.58 13 sin 48.58 sin 1058

A ø 26.58,

C ø 48.58,

c ø 10.1

1 1 7. Area 5 } ab sin C 5 } (10)(14) sin 468 ø 50.4 2 2

c 5 55

Because this is a case of ASA, there is exactly one triangle. 8. A 5 1498,

1 1 8. Area 5 } ac sin B 5 }(19)(8) sin 758 ø 73.4 square units 2 2

9. B 5 348,

b 5 5,

a 5 16

This is the SSA case. Because B is acute: h 5 a sin B 5 16 sin 348 ø 8.9

33.3 square units

239.7 square units

b 5 10

Because A is obtuse and a ≤ b, there is no triangle.

1 1 9. Area 5 }bc sin A 5 }(11)(7) sin 1208 ø 2 2 1 1 10. Area 5 } ab sin C 5 }(20)(24) sin 878 ø 2 2

a 5 7,


square units

Because 8.9 > 5 (h > b), there is no triangle. 10. B 5 708,

b 5 85,

c 5 88


13.5 Exercises (pp. 886 – 888)

Because B is acute: h 5 c sin B 5 88 sin 708 ø 82.7

Skill Practice

Because 82.7 < 85 < 88 (h < a < b), there are two triangles.

1. The law of sines can be used when two angles and the

length of any side are known or when the lengths of two sides and an angle opposite one of the sides is known.

11. C 5 488,

(h) of the triangle is greater than or equal to the side (a) opposite the given angle (A). One triangle: There will be one triangle when the height (h) of the triangle is equal to the side (a) opposite the given angle (A). There will also be one triangle when side (a) opposite the give angle (A) is greater than the other given side. Two triangles: There will be two triangles when the height (h) of the triangle is less than side (a) opposite the given angle (A) and side (a) is less than the other given side (b). a59

b54

c 5 28,

b 5 20


2. No triangle: There will be no triangle when the height

3. A 5 1128,

B 5 648,

This is the ASA case.

c 5 } ø 10.1


a 5 32,


Because C is acute and c > b, there is one triangle. 12.

A 5 1808 2 858 2 478 5 488 b 19 }5} sin 858 sin 488 19 sin 858 sin 488

b 5 } ø 25.5 19 sin 488

c sin 478

}5}

19 sin 478 sin 488

c 5 } ø 18.7 A 5 488,

b ø 25.5,

c ø 18.7

This is the SSA case. Because A is obtuse and a > b, there exists one triangle. 4. A 5 408,

C 5 758,

c 5 20

This is the AAS case. Because this is a case of AAS, there is exactly one triangle.


n2ws13-b.indd 743

743

6/27/06 11:50:09 AM

Chapter 13,

continued

25 16 13. } 5 } ø 0.6210 sin B sin 1048

18. Because A is acute and a > b, there is one triangle. B

16 sin 1048

ø 0.6210 sin B 5 } 25 B ø sin(210.6210) ø 38.48

c

A ø 1808 2 1048 2 38.48 ø 37.68 25 sin 1048

738 A

25 sin 37.68 a 5 } ø 15.7 sin 1048

A ø 37.68, 14.

B ø 38.48,

a ø 15.7

11 sin 738

ø 0.5844 sin B 5 } 18

26 sin 818

B ø sin210.5844 ø 35.88 C ø 1808 2 35.88 2 738 ø 71.28

26 sin 338 sin 818

a 5 } ø 14.3

c 18 }5} sin 71.28 sin 738

26 sin 818

18 sin 71.28 sin 738

}5}

c 5 } ø 17.8

26 sin 668 sin 818

B ø 35.88,

b 5 } ø 24.0 a ø 14.3,

B 5 668, 15.

b ø 24.0

34 sin 738

} 5}

c

a

268

358

A

13

13 sin 268 sin 1198

a 5 } ø 6.5

34 sin 658 sin 738

a ø 23.8,

c sin 1198 5 13 sin 358

}5}

13 sin 358 sin 1198

14 sin 568

c 5 } ø 8.5

ø 0.6827 sin C 5 } 17 A ø 1808 2 568 2 43.18 ø 80.98

20. Because this is an AAS case, there is exactly C

17 sin 80.98 a 5 } ø 20.2 sin 568

C ø 43.18,

c

A

A 5 1808 2 1028 2 438 5 358

43 sin 248 sin 958

}5}

43 sin 958

a 5 } ø 12.3

43 sin 618 sin 958

}5}

a sin 358

21 sin 1028

21 sin 358 sin 1028

}5}

21 sin 1028

b 5 } ø 37.8 a ø 17.6,

b ø 37.8

c sin 438

21 sin 438 sin 1028

c 5 } ø 14.6 A 5 358,

744

1028 B

43 sin 958

a 5 } ø 17.6

C 5 958,

a

a ø 20.2

}5}

b sin 618

438

21

C 5 1808 2 618 2 248 5 958 a sin 248

c ø 8.5

one triangle.

17 a }5} sin 80.98 sin 568

17.

a ø 6.5,

B 5 1198,

C ø sin210.6827 ø 43.18

A ø 80.98,

c sin 358

13 sin 1198

}5}

b ø 32.2

14 sin C

17 sin 568

13 sin 1198

a sin 268

}5}

34 sin 738

b 5 } ø 32.2

16.

C

B 5 1808 2 268 2 35 5 1198

}5}

B 5 658,

c ø 17.8

B

34 sin 428 a 5 } ø 23.8 sin 738 b sin 658

C ø 71.28,

19. Because this is an ASA case, there is exactly one triangle.

B 5 1808 2 738 2 428 5 658 a sin 428

11 sin B

18 sin 738

}5}

b sin 668

C

}5}

B 5 1808 2 818 2 338 5 668 a sin 338

11


a sin 37.68

}5}

18

a ø 12.3,

c ø 14.6


n2ws13-b.indd 744

6/27/06 11:50:17 AM

Chapter 13,

continued

21. Because A is acute:

23. Because A is obtuse and a > b, there is exactly

one triangle.

h 5 b sin A 5 25 sin 388 ø 15.4 Because 15.4 < 19 < 25 (h < a < b), there are two triangles.

C

Triangle 1:

15

10 C

1148

25

A

19

A

c

B

25 19 }5} sin B sin 388

10 sin 1148

B ø sin21 0.6090 ø 37.58

25 sin 388

C ø 1808 2 1148 2 37.58 ø 28.58

B ø sin21 0.8101 ø 54.18 C ø 1808 2 54.18 2 388 ø 87.98

c 15 }5} sin 28.58 sin 1148

c sin 87.98

15 sin 28.58 sin 1148

}5}

c 5 } ø 7.8

25 sin 87.98 sin 54.18

c 5 } ø 30.8 B ø 54.18,

C ø 87.9,

B ø 37.58, c ø 30.8

C a

44 19

498 A

388 c B

328 c

B

C 5 1808 2 498 2 328 5 998

In Triangle 1, B ø 54.18. The obtuse angle B has 54.18 as a reference angle. Copyright © by McDougal Littell, a division of Houghton Mifflin Company.

c ø 7.8

24. Because C is obtuse and c ≤ a, there is no triangle.

C

A

C ø 28.58,

25. Because this is an AAS case, there is exactly one triangle.

Triangle 2:

25

B

ø 0.6090 sin B 5 } 15

ø 0.8101 sin B 5 } 19

25 sin 54.18

c

10 15 }5} sin B sin 1148

388

B ø 1808 2 54.18 ø 125.98 C ø 1808 2 125.98 2 388 ø 16.18

44 sin 498 sin 328

a 5 } ø 62.7 44 sin 328

c 19 }5} sin 16.18 sin 388

c sin 998

}5}

44 sin 998 sin 328

19 sin 16.18 sin 388

c 5 } ø 82.0

c 5 } ø 8.6 B ø 125.98,

a 44 }5} sin 498 sin 328

C ø 16.18,

c ø 8.6

C ø 998,

a ø 62.7,

c ø 82.0

22. Because this is an ASA case, there is exactly one triangle. C

b

a

558 A

648 34

B

C 5 1808 2 558 2 648 5 618 34 a }5} sin 618 sin 558 34 sin 558 sin 618

a 5 } ø 31.8 b sin 648

34 sin 618

}5}

34 sin 648 sin 618

b 5 } ø 34.9 C ø 618,

a ø 31.8,

b ø 34.9


n2ws13-b.indd 745

745

6/27/06 11:50:26 AM

Chapter 13,

continued 1 1 29. Area 5 } ac sin B 5 }(9)(11) sin 1248 ø 41.0 square 2 2

26. Because B is acute:

units

h 5 c sin B 5 32 sin 218 ø 11.5 Because 11.5 < 17 < 32, (h < b < c) there are two triangles. A

Triangle 1: 32

units

218 a

C

32 sin C

17 sin 218

}5}

1 1 32. Area 5 } ab sin C 5 }(25)(17) sin 798 ø 208.6 square 2 2

units

1 1 33. Area 5 } ac sin B 5 }(9)(5)sin 578 ø 18.9 square units 2 2

32 sin 218

ø 0.6746 sin C 5 } 17

1 1 34. Area 5 }ab sin c 5 }(7)(15) sin 968 ø 52.2 square units 2 2

C ø sin 210.6746 ø 42.4 A ø 1808 2 218 2 42.48 ø 116.68 17 sin 218

a sin 116.68

units

1 1 31. Area 5 }bc sin A 5 }(29)(36) sin 348 ø 291.9 square 2 2

17

B

1 1 30. Area 5 } bc sin A 5 }(13)(7) sin 688 ø 42.2 square 2 2

1 1 35. Area 5 } bc sin A 5 }(23)(20) sin 1308 ø 176.2 square 2 2

units

}5}

1 1 36. Area 5 } ac sin B 5 }(19)(14) sin 608 ø 115.2 square 2 2

17 sin 116.68 a 5 } ø 42.4 sin 218

C ø 42.48,

a ø 42.4

Triangle 2:

A

32 218 a

B

1 1 37. Area 5 } ab sin C 5 }(38)(31) sin 298 ø 285.6 square 2 2

units

1 1 38. Area 5 }bc sin A 5 }(15)(13) sin 578 ø 81.8 square 2 2

17

units

C

In Triangle 1, C ø 42.48. The obtuse angle C has 42.48 as a reference angle. C ø 1808 2 42.48 ø 137.68

39.

B 5 1808 2 838 2 598 5 388 24 b }5} sin 598 sin 388 24 sin 388 sin 598

A ø 1808 2 137.68 2 218 ø 21.48

b 5 } ø 17.24

17 a }5} sin 218 sin 21.48

1

A ø 21.48,

C ø 137.68,

a ø 17.3

27. The lengths 5 and 6 were not paired with the correct

opposite angles.

16 20 40. } 5 } sin C sin 668 16 sin 668

ø 0.7308 sin C 5 } 20

sin 558 6

sin C 5

C ø sin210.7308 ø 47.08

}5}

B ø 1808 2 668 2 478 ø 678

5 sin 558

ø 0.6826 sin C 5 } 6 C ø sin 210.6826 ø 43.08

A

678 a

1

1

Area 5 }2 bh, where

B

c

B

c sin 678

}5}

31 sin 678 sin 818

c 5 } ø 28.9 ft

746

1

42. Area formulas:

c

B 5 1808 2 678 2 328 5 818 31 sin 818

1

Area 5 }2 ac sin B 5 }2(29)(24) sin 528 ø 274 square units.

C

328

1

Area 5 }2 ac sin B 5 }2(20)(16) sin 678 ø 147.3 square units. 41. A;

28. B;

31 ft

1

Area 5 }2 bc sin A 5 }2(17.24)(24)(sin 838) ø 205.3 square units.

17 sin 21.48 a 5 } ø 17.3 sin 218

A

b

a

h b

sin A 5 }c . C

c sin A 5 h


A ø 116.68,

units

1

Area 5 }2 bc sin A


n2ws13-b.indd 746

6/27/06 11:50:34 AM

Chapter 13,

continued Problem Solving

B h

43.

a

c

C b

A

b

C

1

h

408

A

Area 5 }2 ah, where }c 5 sin(1808 2 B).

458

300 ft

300 a }5} sin 958 sin 408

h

} c 5 sin B

300 sin 408 sin 958

a 5 } ø 193.6

h 5 c sin B

300 sin 958

b sin 458

1

Area 5 }2 ac sin B

}5}

B

b sin 958 5 300 sin 458

A

300 sin 458 sin 958

a

h

b 5 } ø 212.9

b

One lifeguard is about 193.6 feet away from the windsurfer. The other lifeguard is about 212.9 feet away from the windsurfer.

C

1 Area 5 }2 bh, where 44.

h a

} 5 sin C.

Empire State Building 1458

h 5 a sin C b

1

Statue of Liberty

1 2

} bc sin A 5 } ac sin B


0.6 sin 1458

ø 0.0604 sin A 5 } 5.7

sin B b

}5}

A ø sin210.0604 ø 3.58

1 1 } bc sin A 5 } ab sin C 2 2

B ø 1808 2 1458 2 3.58 5 31.58 5.7 b }5} sin 1458 sin 31.58

c sin A 5 a sin C sin C c

}5}

sin A

sin B

Not drawn to scale

5.7 0.6 }5} sin A sin 1458

b sin A 5 a sin B

sin A a

B Chrysler Building

A

Law of sines:

sin A a

0.6 mi

5.7 mi

Area 5 }2 ab sin C 1 2

B

C 5 1808 2 408 2 458 5 958

Because sin(1808 2 B) 5 sin B:

c

a

5.7 sin 31.58 sin 1458

b 5 } ø 5.2 sin C

5} 5} . So, } c b a

The distance between the Empire State Building and the Statue of Liberty is about 5.2 miles. 45. a.

B

588 C

54 ft 62 ft A

b.

sin 588 sin C }5} 62 54 54 sin 588

ø 0.7386 sin C 5 } 62 C ø 47.68 A ø 1808 2 588 2 47.68 5 74.48 62 a }5} sin 74.48 sin 588 62 sin 74.48 sin 588

a 5 } ø 70.4 ft


n2ws13-b.indd 747

747

6/27/06 11:50:42 AM

Chapter 13,

continued

1 1 c. Area 5 } bc sin A 5 }(62)(54)(sin 74.48) 2 2

ø 1612 ft 2 1612 ft 2 4 200 ft2/bag ø 8.1 bags You will need 9 bags of fertilizer. 1 46. a. Area 5 }(8.5)(8.5) sin 548 ø 29.2. 2

c. 908 maximizes the triangle’s area. The maximum area 1 is 24, which is equal to }2 of the product of the given

side lengths.

49. Because h 5 93 sin 228 ø 34.8, and 34.8 < 36 < 93, there

are two triangles. Triangle 1:

S

The area of nABD is about 29.2 square inches.

93

1 b. Area 5 }(14)(14) sin 328 ø 51.9 2

The area of nBCD is about 51.9 square inches. c. The area of the kite is about 29.2 1 51.9 5

81.1 square inches. 47. a.

36

228 E

s

M

sin 228 sin M }5} 36 93 93 sin 228

sin M 5 } 36 M ø 75.48

h

S ø 1808 2 75.48 2 228 5 82.68 36 sin 228

s sin 82.68

}5}

728

300 m

36 sin 82.68 sin 228

s 5 } ø 95 S

Triangle 2:

300 tan 638 5 } b

93

300 b 5 } ø 152.9 tan 638

The person is about 152.9 meters away from the base of the cliff. b. You could use the definition of tangent to write the 300 1 h and solve for h. Or, you equation tan 728 5 } 152.9

E

s

S ø 1808 2 104.68 2 228 5 53.48 s sin 53.48

152.9 sin(1808 2 728 2 908)

300 1 h sin 728

152.9 sin 188

}5}

152.9 sin 728 sin 188

h 5 } 2 300 ø 170.6 The height of the building is about 170.6 meters. 48. a. Sample answer:

36 sin 53.48 sin 228

s 5 } ø 77 The possible distances between Mercury and Earth are about 95 million miles or about 77 million miles. Mixed Review }

}

}

50. 6Ï 13 2 Ï 13 5 5Ï 13

51. 5(250)1/3 2 10(54)1/3 5 25(2)1/3 2 30(2)1/3 5 25(2)1/3 4}

4}

4}

4}

4}

52. 22 Ï 160 1 4 Ï 810 5 24 Ï 10 1 12 Ï 10 5 8 Ï 10 53. 5(20)1/2 2 3(45)1/2 5 10(5)1/2 2 9(5)1/2 5 51/2 3}

6

36 sin 228

}5}

Using method 2: 300 1 h sin 728

M

In triangle 1, M ø 75.48. The obtuse angle M has 75.48 as a reference angle, so M ø 1808 2 75.48 5 104.68.

could use the law of sines to write an equation in terms of h.

} 5 }}

36

228

3}

3}

3}

3}

54. 9 Ï 56 2 Ï 189 5 18 Ï 7 2 3 Ï 7 5 15Ï 7

x 8

1 2

Area 5 }(6)(8) sin x 5 24 sin x An equation is y 5 24 sin x, where the length of the sides of the triangle are 6 and 8. b. No; The area increases as x increases up to 908. At

x 5 908, the area reaches a maximum and then begins to decrease.

55. 26(88)1/3 1 9(297)1/3 5 212(11)1/3 1 27(11)1/3

5 15(11)1/3 xy 5x 2 x2 5x 5 1 56. } p}5} p} 5 }2 y 20y 4y x 3y 20xy 2 10xy 20x 2y 2 20x2y2 xz3 40x 3y 3 10 57. } 4 } 5} 4 }3 5 } p} 5 2x3y 2z2 z z 2xyz 10 x 2yz 3 xz


638 b

3(x 2 2 3) x17 3x 2 2 9 x17 x17 58. } p } 5} p} 5} x21 2(x 2 1) x21 6x2 2 18 6(x 2 2 3) 4(x 1 2)(x 2 2) x12 4x 2 2 16 x 1 5 x15 59. } p } 5 }} p}5} x25 (x 1 5)(x 2 5) 4(x 2 2) x 2 2 25 4x 2 8

748


n2ws13-b.indd 748

6/27/06 11:50:50 AM

Chapter 13,

continued

(x 1 3)(x 1 2) x 2 1 5x 1 6 1 60. }} 4 (x 1 3) 5 }} p} (3x 1 7)(x 1 2) x 1 3 3x 2 1 13x 1 14

67. u 5 cos210.47 ø 62.08 y

1

5} 3x 1 7 u

(3x 1 4)(2x 1 1) 6x 2 1 11x 1 4 6x 2 8 x11 61. }} 4} 5 }} p} x11 2(3x 2 4) 2x 2 1 3x 2 1 2x 2 1 3x 2 1 (3x 1 4)(2x 1 1)(x 1 1) 5 }} 2(3x 2 4)(2x2 1 3x 2 1)

628 628

62. u 5 cos210.75 ø 41.48

x

u ø 3608 2 62.08 ø 298.08

y

Lesson 13.6 u

41.48 41.48 x

13.6 Guided Practice (p. 891) 1.

b2 5 a2 1 c2 2 20 c cos B 5 82 1 102 2 2(8)(10) cos 488 ø 56.93 b ø 7.55

u ø 3608 2 41.48 ø 318.68

sin C 10

y

53.18

u

sin 488 7.55

}5}

63. u 5 cos21 (20.6) ø 126.98

10 sin 488

ø 0.9843 sin C 5 } 7.55

126.98

C ø sin210.9843 ø 79.88 x

53.18

A ø 1808 2 488 2 79.88 5 52.28 A ø 52.28, C ø 79.88, b ø 7.55

u ø 1808 1 53.18 ø 233.18

2.

16 2 5 14 2 1 9 2 2 2(14)(9) cos B

64. u 5 cos21(20.35) ø 110.58

256 5 277 2 252 cos B

y

0.0833 ø cos B B ø cos21 0.0833 ø 85.28

110.58


b 2 5 a 2 1 c 2 2 2ac cos B

sin A 14

x

sin 85.28 16

}5}

14 sin 85.28

ø 0.8719 sin A 5 } 16 65. u 5 cos210.92 ø 23.18

A ø sin21 0.8719 ø 60.78

y

u

C ø 1808 2 85.28 2 60.78 5 34.18 23.18 23.18 x

A ø 60.78, B ø 85.28, C ø 34.18 3.

b 2 5 a 2 1 c 2 2 2ac cos B 3352 5 1932 1 1862 2 2(193)(186) cos B 112,225 5 71,845 2 71,796 cos B

u ø 3608 2 23.18 ø 336.98 66. u 5 cos21 (20.28) ø 106.38

73.78

u

B 5 cos21(20.5624) ø 124.28 The step angle u is about 124.28.

y

73.78

20.5624 5 cos B

1 1 4. s 5 }(a 1 b 1 c) 5 }(8 1 11 1 5) 5 12 2 2

106.38 x

}}

Area 5 Ïs(s 2 a)(s 2 b)(s 2 c)

}}}

5 Ï12(12 2 8)(12 2 11)(12 2 5) }

5 Ï336 ø 18.3 square units

u ø 1808 1 73.78 ø 253.78

1 1 5. s 5 }(a 1 b 1 c) 5 }(4 1 9 1 7) 5 10 2 2 }}


}}}

}

5 Ï10(10 2 4)(10 2 9)(10 2 7) 5 Ï 180 ø 13.4 square units


n2ws13-b.indd 749

749

6/27/06 11:50:59 AM

Chapter 13, 1

1

s 5 }2 (a 1 b 1 c) 5 }2 (15 1 23 1 12) 5 25

11. b 2 5 a 2 1 c 2 2 2ac cos B

5 82 1 62 2 2(8)(6) cos 258

}}


}}}

ø 12.99

}

5 Ï25(25 2 15)(25 2 23)(25 2 12) 5 Ï6500

b ø 3.60

ø 80.6 square units

6 sin 258

ø 0.7044 sin C 5 } 3.60

Skill Practice 1. In a triangle with sides of length a, b, and c, 1 } (a 1 b 1 c) is called the semiperimeter. 2 2. Heron’s formula states that the area of a triangle is equal

to the square root of the product of the semiperimeter and the differences between the semiperimeter and each of the sides of the triangle. 3. Law of cosines

4. Law of sines

5. Law of sines

6. Law of cosines

C ø sin21 0.7044 ø 44.88 A ø 1808 2 258 2 44.88 5 110.28 A ø 110.28, C ø 44.88, b ø 3.60 12. a 2 5 b 2 1 c 2 2 2bc cos A

5 152 1 242 2 2(15)(24)cos 1038 ø 962.96 a ø 31.03 sin 1038 31.03

7. Law of sines

15 sin 1038

ø 0.4710 sin B 5 } 31.03

5 72 1 92 2 2(7)(9) cos 358 ø 26.79 a ø 5.18

B ø sin210.4718 ø 28.18

sin 358 sin B }5} 5.18 7

C ø 1808 2 1038 2 28.18 5 48.98 B ø 28.18, C ø 48.98, a ø 31.03

7 sin 358

ø 0.7751 sin B 5 } 5.18

13.

b 2 5 a 2 1 c 2 2 2ac cos B

B ø sin21 0.7751 ø 50.88

282 5 182 1 132 2 2(18)(13) cos B

C ø 1808 2 358 2 50.88 5 94.28

784 5 493 2 468 cos B

B ø 50.88, C ø 94.28, a ø 5.18

20.6218 ø cos B

9. a 2 5 b 2 1 c 2 2 2bc cos A

B ø cos21(20.6218) ø 128.48

5 232 1 262 2 2(23)(26) cos 1148 ø 1691.46

sin A 18

18 sin 128.48

sin C 26

sin 1148 41.1

ø 0.5038 sin A 5 } 28

}5}

A ø sin 21 0.5038 ø 30.38

26 sin 1148

ø 0.5779 sin C 5 } 41.1 C ø sin

21

B ø 30.78; C ø 35.38, a ø 41.1 2

2

a 5 b 1 c 2 2bc cos A 2

2

2

25 5 23 1 14 2 2(23)(14) cos A 625 5 725 2 644 cos A 0.1553 ø cos A A ø cos

21

0.1553 ø 81.18

23 sin 81.18 ø 0.9089 sin B 5 } 25

B ø sin

A ø 30.38, B ø 128.48, C ø 21.38 14.

a 2 5 b 2 1 c 2 2 2bc cos A 382 5 312 1 352 2 2(31)(35) cos A 1444 5 2186 2 2170 cos A 2742 5 22170 cos A 0.3419 ø cos A A ø cos21 0.3419 ø 70.08

sin B sin 81.18 }5} 23 25

21

C ø 1808 2 128.48 2 30.38 ø 21.38

0.5779 ø 35.38

B ø 1808 2 1148 2 35.38 5 30.78 2

sin 128.48 28

}5}

a ø 41.1

0.9089 ø 65.48

C ø 1808 2 81.18 2 65.48 5 33.58 A ø 81.18, B ø 65.48, C ø 33.58

750

sin B 15

}5}

8. a2 5 b2 1 c2 2 2 bc cos A

10.

sin 258 3.60

sin C 6

}5}

13.6 Exercises (pp. 892–894)

sin A a

sin B b

sin 70.08 38

sin B 31

}5} }5}

31 sin 70.08

ø 0.7666 sin B 5 } 38


6.

continued

B ø sin21 0.7666 ø 50.08 C ø 1808 2 70.08 2 50.08 ø 60.08 A ø 70.08, B ø 50.08, C ø 60.08


n2ws13-b.indd 750

6/27/06 11:51:04 AM

Chapter 13,

continued

15. c 2 5 a 2 1 b 2 2 2ab cos C 2

19. c 2 5 a 2 1 b 2 2 2 ab cos C

2

5 35 2 1 432 2 2(35)(43) cos 968 ø 3388.63

5 17 1 20 2 2(17)(20) cos 488 ø 233.99

c ø 58.2

c ø 15.3

sin 968 sin A }5} 58.2 35

sin 488 15.3

sin A 17

}5}

35 sin 968

ø 0.5981 sin A 5 } 58.2

17 sin 488

ø 0.8257 sin A 5 } 15.3

A ø sin210.5981 ø 36.78

A ø sin21 0.8257 ø 55.78

B ø 1808 2 968 2 36.78 5 47.38

B ø 1808 2 488 2 55.78 5 76.38

A ø 36.78, B ø 47.38, c ø 58.2

A ø 55.78, B ø 76.38, c ø 15.3 2

2

20. D;

2

292 5 172 1 142 2 2(17)(14)cos B

16. b 5 a 1 c 2 2ac cos B

5 292 1 382 2 2(29)(38) cos 638

841 5 485 2 476 cos B

ø 1284.40

20.7479 ø cos B

b ø 35.8

B ø cos21 (2 0.7479) ø 138.48

sin 638 35.8

sin C 29

}5}

21.

C ø sin

A ø 46.08, C ø 71.08, b ø 35.8 17.

}}}}

5 Ï25.5(25.5 2 19)(25.5 2 21)(25.5 2 11) ø 104 units2

22.

}}}

5 Ï31(31 2 20)(31 2 30)(31 2 12) ø 80.5 units2

23.


}}}

5 Ï77(77 2 47)(77 2 58)(77 2 49)

sin 125.78 12

10 sin 125.78

ø 1108.6 units2 24.

}}

5 Ï13.5(13.5 2 12)(13.5 2 7)(13.5 2 8) ø 26.9 units2

A ø 42.68, B ø 11.78, C ø 125.78 242 5 232 1 202 2 2(23)(20) cos B

25.

}}}

5 Ï13(13 2 5)(13 2 11)(13 2 10) ø 25.0 units2

26.

sin 67.48 24

}}}

5 Ï34(34 2 25)(34 2 24)(34 2 19)

23 sin 67.48 24

C ø 1808 2 67.48 2 62.28 5 50.48 A ø 62.28, Bø 67.48, C ø 50.48

1

}}

sin A 5 } ø 0.8847 A ø sin 0.8847 ø 62.28

1

s 5 }2 (a 1 b 1 c) 5 }2 (25 1 24 1 19) 5 34 Area 5 Ïs(s 2 a)(s 2 b)(s 2 c)

}5}

21

1

}}

0.3837 ø cos B sin A 23

1


576 5 929 2 920 cos B B ø cos210.3837 ø 67.48

1

}}}}

B ø 1808 2 125.78 2 42.68 5 11.7 b 2 5 a 2 1 c 2 2 2ac cos B

1

s 5 }2 (a 1 b 1 c) 5 }2 (12 1 7 1 8) 5 13.5 Area 5 Ïs(s 2 a)(s 2 b)(s 2 c)

A ø sin21 0.6767 ø 42.68

18.

1

}}

}5}

ø 0.6767 sin A 5 } 12

1


C ø cos21(20.5833) ø 125.78 sin A 10

1

}}

122 5 102 1 32 2 2(10)(3) cos C 144 5 109 2 60 cos C

1


c 2 5 a 2 1 b 2 2 2 ab cos C

20.5833 ø cos C

1

}}

0.9458 ø 71.08

A ø 1808 2 638 2 71.08 ø 46.08

1

s 5 }2 (a 1 b 1 c) 5 }2 (19 1 21 1 11) 5 25.5 Area 5 Ïs(s 2 a)(s 2 b)(s 2 c)

38 sin 638

ø 0.9458 sin C 5 } 35.8 21

b 2 5 a 2 1 c 2 2 2ac cos B

ø 214.2 units2 27.

1

1

s 5 }2 (a 1 b 1 c) 5 }2 (14 1 20 1 28) 5 31 }}


}}}

5 Ï31(31 2 14)(31 2 20)(31 2 28) ø 131.9 units2


n2ws13-b.indd 751

751

6/27/06 11:51:08 AM

Chapter 13, 1

1 2

s 5 }2 (a 1 b 1 c) 5 }2 (31 1 23 1 17) 5 35.5

} 5 cos B

1

}}


B 5 cos21 }2 5 608

}}}}

5 Ï35.5(35.5 2 31)(35.5 2 23)(35.5 2 17)

C 5 1808 2 608 2 608 5 608

ø 192.2 units2 1

29.

Because an equilateral triangle has three sides of the same length x, the law of cosines shows that each angle in the equilateral triangle is 608.

1

s 5 }2 (a 1 b 1 c) 5 }2 (81 1 67 1 71) 5 109.5 }}


}}}}

5 Ï109.5(109.5 2 81)(109.5 2 67)(109.5 2 71)

35. A form of the law of cosines equation is:

b 2 1 c 2 2 a2

cos A 5 } 2bc

2

ø 2259.7 units 1

30.

An error was made when substituting values of a, b, and c into this equation. The value of 2bc is 2(15)(10), not 2(18)(15).

1

s 5 }2 (a 1 b 1 c) 5 }2 (43 1 59 1 48) 5 75 }}


152 1 102 2 182

cos A 5 }} ø 0.0033 2(15)(10)

}}}

5 Ï75(75 2 43)(75 2 59)(75 2 48) ø 1018.2 units2 1

31.

A ø cos210.0033 ø 89.88 1

s 5 }2 (a 1 b 1 c) 5 }2 (51 1 51 1 43) 5 72.5

36.

B

}}


5 Ï72.5(72.5 2 51)(72.5 2 51)(72.5 2 43) 5 994.3 units2

14 a }5} sin 448 sin 728

ø 234.6 units2

14 sin 728 sin 448

1 1 33. B; s 5 }(a 1 b 1 c) 5 }(21 1 16 1 13) 5 25 2 2

a 5 } ø 19.2

}}


14 sin 448

5 Ï25(25 2 21)(25 2 16)(25 2 13)

14 sin 648 sin 448

c 5 } ø 18.1

ø 104 units2 B

C ø 648, a ø 19.2, c ø 18.1 c

37.

x

B c

x 2

C 2

2

A

988

b

18 378

C

a 5 b 1 c 2 2bc cos A

ASA: one triangle

x 2 5 x 2 1 x 2 2 2(x)(x) cos A

A 5 1808 2 988 2 378 5 458

x 2 5 2x 2 2 2x 2 cos A 2x 2 5 22x 2 cos A

18 sin 458

b sin 988

}5}

18 sin 988 sin 458

1 2

b 5 } ø 25.2

} 5 cos A

1

A 5 cos21 }2 5 608 b 2 5 a 2 1 c 2 2 2 ac cos B x 2 5 x 2 1 x 2 2 2(x)(x)cos B 2x 2 5 22x 2 cos B

752

c sin 648

}5}

}}}

A

C

C 5 1808 2 728 2 448 5 648

}}}

5 Ï62(62 2 38)(62 2 25)(62 2 61)

x

14

AAS: One triangle

}}


34.

a

728 A

1 1 s 5 }2 (a 1 b 1 c) 5 }2 (38 1 25 1 61) 5 62

32.

448

c

}}}}

18 sin 458

c sin 378

}5}

18 sin 378 sin 458

c 5 } ø 15.3 A ø 458, b ø 25.2, c ø 15.3


1

28.

continued


n2ws13-b.indd 752

6/27/06 11:51:16 AM

Chapter 13, 38.

continued Triangle 2:

B c

A

A

36

12 658

21

C

C

From Triangle 1: B ø 1808 2 59.08 5 121.08

SAS: one triangle c 2 5 a 2 1 b 2 2 2ab cos C 2

A ø 1808 2 121.08 2 408 5 19.0

2

5 12 1 21 2 2(12)(21) cos 658 ø 372.0

27 a }5} sin 408 sin 198

c ø 19.3

27 sin 198 sin 408

a 5 } ø 13.7

sin 658 19.3

sin A 12

}5}

A ø 19.08, B ø 121.08, a ø 13.7

12 sin 658

ø 0.5635 sin A 5 } 19.3

41. SSS: One Triangle

A ø sin210.5635 ø 34.38

A

B ø 1808 2 658 2 34.38 5 80.78

27

A ø 34.38, B ø 80.78, C ø 19.3

B

39. A

34

C

a 5 b 1 c2 2 2bc cos A

b

B

15

1156 5 1090 2 1026 cos A

C

}

20.0643 ø cos A }

A ø cos21(20.0643) ø 93.78

}

b 5 Ïa 1 c 5 Ï15 1 6 5 Ï261 ø 16.2 2

2

2

2

}5}

}5}

19 sin 93.78

15 sin 908

ø 0.5577 sin B ø } 34

ø 0.9259 sin A 5 } 16.2

B ø sin21 0.5577 ø 33.98

A 5 sin21 0.9259 ø 67.88

C ø 1808 2 93.78 2 33.98 5 52.48

C ø 1808 2 908 2 67.88 5 22.28

A ø 93.78, B ø 33.98, C ø 52.48

A ø 67.88, C ø 22.28, b ø 16.2 40. C is acute: h 5 b sin C 5 36 sin 408 ø 23.1

42.

C

Because C is acute and 23.1 < 27 < 36 (h < c < b), there are two triangles. Triangle 1: 36 408

b A

A

C

sin B 19

sin 93.78 34

sin 908 16.2

sin A 15

2

342 5 192 1 272 2 2(19)(27)cos A

SAS: one triangle


19

2

6

27

408 a B

a c2x

B

x cos A 5 }b → x 5 b cos A

27 a

x

h

B

27 36 }5} sin B sin 408 36 sin 408

ø 0.8571 sin B 5 } 27

x2 1 h2 5 b2 → h 2 5 b 2 2 x2 a 2 5 h 2 1 (c 2 x) 2 a 2 5 h 2 1 c 2 2 2cx 1 x 2 a 2 5 b 2 2 x 2 1 c 2 2 2cx 1 x 2 a 2 5 b 2 1 c 2 2 2cx

B ø sin 0.8571 ø 59.08

a 2 5 b 2 1 c 2 2 2c(b cos A)

A ø 1808 2 59.08 2 408 5 81.08

a 2 5 b 2 1 c 2 2 2bc cos A

21

a sin 81.08

27 sin 408

The other two versions can be derived in a similar fashion.

27 sin 81.08 sin 408

If an angle of n ABC is a right angle, then its cosine is equal to zero and the law of cosines reduces to the Pythagorean Theorem.

}5}

a 5 } ø 41.5 A ø 81.08, B ø 59.08, a ø 41.5


n2ws13-b.indd 753

753

6/27/06 11:51:25 AM

Chapter 13,

continued

Problem Solving 26 ft 17 ft

u

608

After 1 hour, a 5 3 miles and b 5 2 miles. c 2 5 32 1 22 2 2(3)(2) cos 608 5 7 }

c ø Ï7 ø 2.6

u ø cos21(20.4932) ø 119.68 The two flyers will meet at an angle of about 119.68. 1

3 mi

C

20.4932 ø cos u

1

The hikers are about 2.6 miles apart after 1 hour. b. After t hours, a 5 3t and b 5 2t.

s 5 }2 (a 1 b 1 c) 5 }2 (18 1 23 1 8) 5 24.5

c 2 5 a 2 1 b 2 2 2ab cos 608

}}

}}}}

100 5 9t 2 1 4t 2 2 6t 2

5 Ï24.5(24.5 2 18)(24.5 2 23)(24.5 2 8)

100 5 7t 2

ø 62.8

Ï

The area of the research triangle is about 62.8 square miles.

ø 3 hours 47 minutes

1

c.

Ï1716.5(1716.5 2 743)(1716.5 2 1210)(1716.5 2 1480) ø 447,399.18

100

The hikers are out of range of each other after about 3 hours and 47 minutes.

s 5 }2 (743 1 1210 1 1480) 5 1716.5 }}}}}

}

tø } ø 3.78 7

45. Area of n ABD:

Area 5

A

c

u B

4 mi 608

Area of n BCD:

1

10 2 5 (3t) 2 1 (2t)2 2 2(3t)(2t)}2


6 mi

C

1

s 5 }2 (858 1 1000 1 1480) 5 1669 }}}}

Area 5 Ï1669(1669 2 858)(1669 2 1000)(1669 2 1480) ø 413,697.18 Total area ø 447,399.18 1 413,697.18 ø 861,096.36 1 acre 43,560 ft

861,096.36 ft2 p }2 ø 19.8 acres The area of the land is about 19.8 acres. You could also find the area by finding AC, using Heron’s forumla to find the areas of n ACD and n ACB, then adding them together and converting the area from square feet to acres. 46. a. x 2 5 260 2 1 400 2 2 2(260)(400) cos 158

ø 26,687.43

2

c 5 62 1 42 2 2(6)(4)cos 608 5 28 }

c 5 Ï28 ø 5.3 Because the second hiker has to travel 5.3 miles: d 5 rt 5.3 5 3t 5.3

t5} ø 1.77 ø 1 hour 46 minutes 3 It will take the second hiker about 1 hour and 46 minutes to meet the first hiker. Using law of sines: sin 608 5.3 6 sin 608 } sin A 5 5.3 ø 0.9804 sin A 6

}5}

A 5 sin21 0.9804 ø 78.68

x ø 163.36 The distance from the golfer’s ball to the hole is about 163.36 yards. 10 2 5 163.36 2 1 163.362 2 2(163.36)(163.36) cos u 100 ø 53.372.98 2 53.372.98 cos u 0.9981 ø cos u

u ø cos21 (0.9981) ø 3.538 The maximum angle u by which the ball can be off target in order to land no more than 10 yards from the hole is about 3.538.

754

B

13 ft

676 5 458 2 442 cos u

b.

c

2 mi

26 2 5 172 1 132 2 2(17)(13) cos u

44.

A

Because alternate interior angles are congruent, A 5 u, so u ø 78.68. The second hiker should walk about 78.68 west of north.


43.

47.a.


n2ws13-b.indd 754

6/27/06 11:51:32 AM

Chapter 13, 48.

continued 56.

B N 470 mi

558 W

u

E

A

3

1

29

26

217

3

218

24

26

8

7

a C

890 mi

1

78

7 x 3 2 9x2 1 26x 2 17 }} 5 x 2 2 6x 1 8 1 } x23 x23

S

u 5 908 2 558 2 78 5 288 a 2 5 4702 1 8902 2 2(470)(890) cos 288 ø 274,326.04

57.

1

22

a ø 523.76 The distance between cities B and C is about 523.76 miles.

1

50.

y

58.

y

5

1

y 5 4x 2 5

1

y 5 9 2 4x

x

21

4

22

2

210

6

21

5

23

10

1

231

8

222

5

30

25

15

6

21

3

27

x 4 1 x 3 2 31x 2 1 8x 2 22 x25

7 x25

59. f 1 g 5 x 1 7 1 10x 5 11x 1 7

Domain: all real numbers

x

21

7

}} 5 x 3 1 6x 2 2 x 1 3 2 }

1

1

3

10 x 4 1 x 3 1 3x 2 1 7x 1 4 }} 5 x 3 2 x 2 1 5x 2 3 1 } x12 x12

Mixed Review 49.

1

60. f 2 g 5 5x1/2 2 (24x1/2) 5 9x1/2

210 b 51. vertex: x 5 2} 5 2} 5 5 2a 2(1)

Domain: x ≥ 0

y 5 52 2 10(5) 1 12 5 213

61. f p g 5 (3x 2 2)(2x3) 5 6x 4 2 4x 3

(5, 213)

Domain: all real numbers

f(x)

62. f ( g(x)) 5 f(8x 1 11) 5 (8x 1 11)2 2 6


5 64x2 1 176x 1 115 Domain: all real numbers

4 x

22

Quiz 13.5–13.6 (p. 894) 1.

f(x) 5 x 2 2 10x 1 12

52.

53.

y

B 12

y

y52 x1723

a

508 1

21

748

y5 x

C

ASA: one triangle

1 21

b

A

1.5e 2x

x

C 5 1808 2 748 2 508 5 568 a sin 508

12 sin 568

}5}

54.

y

12 sin 508 sin 568

a 5 } ø 11.1

1 x

21

b sin 748

12 sin 568

}5}

g(x) 5 ln (x 1 4)

12 sin 748 sin 568

b 5 } ø 13.9 55.

24

1

1

8

17

1

24

216

24

4

1

23

C 5 568, a ø 11.1, b ø 13.9

3 x 3 1 8x 2 1 17x 1 1 }} 5 x 2 1 4x 1 1 2 } x14 x14


n2ws13-b.indd 755

755

6/27/06 11:51:44 AM

Chapter 13,

continued

2. h 5 a sin C 5 18 sin 668 ø 16.4

4. SAS: one triangle

Because 16.4 < 17 < 18, there are two triangles. Triangle 1:

B

B

18

c

A

17

34

c 2 5 262 1 342 2 2(26)(34) cos 1188 ø 2662.03

668 b

C

sin A 18

c ø 51.6

A

sin 1188 51.6

sin A 26

sin 668 17

}5}

18 sin 668

ø 0.4449 sin A 5 } 51.6

}5}

ø 0.9673 sin A 5 } 17 A ø sin21 0.9673 ø 75.38 B ø 1808 2 668 2 75.38 5 38.78 17 sin 668

b sin 38.78

26 1188 C

}5}

26 sin 1188

A ø sin21 0.4449 ø 26.48 B ø 1808 2 1188 2 26.48 ø 35.68 A ø 26.48, B ø 35.68, c ø 51.6 5. AAS: one triangle

17 sin 38.78 b 5 } ø 11.6 sin 668

A ø 75.38, B ø 38.78, b ø 11.6 Triangle 2:

B 31

c 1028 A

B

258 b

C

B 5 1808 2 1028 2 258 5 538 18

17

31 b }5} sin 1028 sin 538 31 sin 538 sin 1028

b 5 } ø 25.3 C b A

A ø 1808 2 75.38 5 104.78

c sin 258

31 sin 258 sin 1028

c 5 } ø 13.4

B ø 1808 2 668 2 104.78 5 9.38

b sin 9.38

17 sin 668

}5}

B 5 538, b ø 25.3, c ø 13.4

17 sin 9.38 sin 668

b 5 } ø 3.0 A ø 104.78, B ø 9.38, b ø 3.0 3. SSS: one triangle B

A

6. SSS: one triangle B 49

38 A

52 2

23

2

2

1444 5 5105 2 5096 cos C 0.7184 ø cos C

C 2

2

23 5 20 1 14 2 2(14)(20) cos C 529 5 596 2 560 cos C 0.1196 ø cos C C ø cos21 0.1196 ø 83.18 sin A 20

C

38 5 49 1 522 2 2(49)(52) cos C

20

14

31 sin 1028

}5}

sin 83.18 23

}5}

20 sin 83.18

ø 0.8633 sin A 5 } 23

C ø cos21 0.7184 ø 44.18 sin 44.1 38

sin B 52

}5}

52 sin 44.18

ø 0.9523 sin B 5 } 38 B ø sin21 0.9523 ø 72.28 A ø 1808 2 44.18 2 72.28 5 63.78


668

A ø 63.78, B ø 72.28, C ø 44.18

A ø sin21 0.8633 ø 59.78 B ø 1808 2 83.18 2 59.7 5 37.28 A ø 59.78, B ø 37.28, C ø 83.18

756


n2ws13-b.indd 756

6/27/06 11:51:53 AM

Chapter 13,

continued

7. SAS: one triangle

15. Area of base:

B

1

s 5 }2 (8 1 10 1 13) 5 15.5

538

}}}}

Area 5 Ï15.5(15.5 2 8)(15.5 2 10)(15.5 2 13)

41

29

ø 39.98 A

b

V 5 Area of base p height ø 39.98 p 5 ø 199.9

C

The volume of the prism is about 199.9 cubic centimeters.

b2 5 292 1 412 2 2(29)(41) cos 538 ø 1090.88 b ø 33.0

Problem Solving Workshop 13.6 (p. 895)

sin 538 33.0

sin C 29

}5}

1.

C

29 sin 538 ø 0.7018 sin C 5 } 33.0

12 B

C ø sin21 0.7018 ø 44.68 A ø 1808 2 538 2 44.68 5 82.48

26

26 2 x

A

h 2 1 (26 2 x)2 5 17 2

8. ASA: one triangle

122 2 x 2 1 (26 2 x)2 5 172

B

144 2 x 2 1 676 2 52x 1 x 2 5 289

488

A

x

h 2 1 x 2 5 12 2 → h 2 5 122 2 x 2

A ø 82.48, C ø 44.68, b ø 33.0

5

17

h

a

1128

252x 5 2531 b

x ø 10.2

C

}

h ø Ï122 2 10.22 ø 6.3

C 5 1808 2 1128 2 488 5 208

1

5 sin 1128 sin 208

a 5 } ø 13.6

The area is about 81.9 square units. 2.

B

b sin 488

5 sin 208


}5}

87

5 sin 488 sin 208

b 5 } ø 10.9 C 5 208, a ø 13.6, b ø 10.9 1 1 9. Area 5 } ac sin B 5 }(13)(15) sin 948 ø 97.3 units2 2 2 1 1 10. Area 5 } ab sin C 5 }(16)(11) sin 188 ø 27.2 units2 2 2 11.

1 s 5 }2 (18 1 25 1 19) 5 31

A

h 63

x

92 2 x

92

C

h 2 1 x 2 5 87 2 → h 2 5 872 2 x 2 h 2 1 (92 2 x)2 5 632 2

87 2 x 2 1 (92 2 x)2 5 632 7569 2 x 2 1 8464 2 184x 1 x 2 5 3969 2184x 5 212,064 x ø 65.6

}}}

Area 5 Ï31(31 2 18)(31 2 25)(31 2 19) 2

ø 170.3 units 12.

1

s 5 }2 (27 1 21 1 37) 5 42.5 }}}}

Area 5 Ï42.5(42.5 2 27)(42.5 2 21)(42.5 2 37)

1 2

A 5 }2 bh ø }(26)(6.3) ø 81.9

a sin 1128

5 sin 208

}5}

}}

h ø Ï87 2 (65.6) ø 57.1 2

1

2

1 2

A 5 }2 bh ø }(92)(57.1) 5 2626.6 The area is about 2626.6 square units.

2

ø 279.1 units 13.

1

s 5 }2 (62 1 47 1 53) 5 81 }}}

Area 5 Ï81(81 2 62)(81 2 47)(81 2 53) ø 1210.4 units2

1 1 14. Area 5 } bc sin A 5 } (44)(36) sin 708 ø 744.2 units2 2 2


n2ws13-b.indd 757

757

6/27/06 11:52:04 AM

Chapter 13, 3.

continued Mixed Review of Problem Solving (p. 896)

C 101

B

x

153

1.

94

h

153 2 x

d

540 yds

A

h 2 1 x 2 5 1012 → h 2 5 1012 2 x 2 2

2

1008 330 yds

2

h 1 (153 2 x) 5 94

1012 2 x 2 1 (153 2 x)2 5 942

a. d 2 5 5402 1 3302 2 2(540)(330) cos 1008

ø 462,388.21

10,201 2 x 2 1 23,409 2 306x 1 x 2 5 8836

d ø 680

33,610 2 306x 5 8836

The third side is about 680 yards.

2306x 5 224,774 x ø 80.96

}}

h ø Ï101 2 (80.96) ø 60.39 2

2

b.

}}}}

Area 5 Ï775(775 2 540)(775 2 330)(775 2 680)

1 1 A 5 }2 bh ø }(153)(60.39) ø 4619.8 2

ø 87,746

1

The area is about 4619.8 square units. 4.

21

$2500

2

The land costs about $45,323.

200

240

h

x

350 2 x

2. a.

u1 2

A

423

h 1 x 5 200 → h 5 2002 2 x 2 2

1 acre

87,746 yd2 }2 } ø $45,323 4840 yd 1 acre

C

B

1

s 5 }2 (540 1 330 1 680) 5 775

2

2

2

h 2 1 (350 2 x)2 5 2402 2002 2 x 2 1 (350 2 x)2 5 2402

206.5

40,000 2 x 2 1 122,500 2 700x 1 x 2 5 57,600 u1

2700x 5 2104,900

206.5

206.5

u1 2

} ø tan210.4882 ø 26.08

}}

h ø Ï2002 2 149.92 ø 132.4

u1 ø 528

1

A 5 }2 bh 5 }2 (350)(132.4) ø 23,170 The area is about 23,170 square yards.

b.

u2 2

5. 43

h x

50

423

38

50 2 x

522.5

h 2 1 x 2 5 432 → h 2 5 43 2 2 x 2 h 2 1 (50 2 x)2 5 382 43 2 2 x 2 1 (50 2 x)2 5 382 2

1849 2 x 1 2500 2 100x 1 x 2 5 1444 4349 2 100x 5 1444 2100x 5 22905 }}

x 5 29.05

h 5 Ï432 2 29.052 ø 31.7 1

1 2

A 5 }2 bh ø }(50)(31.7) ø 792.5 The area is about 792.5 square feet.

u2

1045

522.5

522.5

tan } 5} ø 1.2352 423 2 u2 2

} ø tan21 1.2352 ø 51.08

u2 ø 1028 1 3. Area 5 }ab sin C 2 1

40 5 }2 ab sin 308 1

40 5 }4 ab


tan } 5} ø 0.4882 423 2

x ø 149.9 1

413

160 5 ab Sample answer: a 5 10, b 5 16 The lengths of the two sides could be 10 inches and 16 inches.

758


n2ws13-b.indd 758

6/27/06 11:52:14 AM

Chapter 13,

continued

3 4. tanu 5 } 20

6.

B

1 2

3 u 5 tan21 } ø 8.58 20

c

The angle u at which the beach slopes is about 8.58.

A

sin 258 sin G 5. } 5 } 95 60.5

4 C

5

}

}

c 5 Ï42 1 52 5 Ï41 5 Ï 41

5Ï41

adj

4 Ï 41

4Ï41 41

opp

5

hyp

Ï 41

hyp

Ï 41

adj

4

sin B 5 } 5} } 5 } 41 hyp

G ø sin21 0.2691 ø 15.68

5} cos B 5 } } 5 } hyp

P ø 1808 2 258 2 15.68 5 139.48 ø 21,413.07

}

csc B 5 } opp 5 } 5

}

d ø Ï21,413.07 ø 146.33 You can find angle G using the law of sines, and angle P using the Triangle Sum Theorem. Then you can find d using the law of cosines. An outfielder at G must throw the ball about 146.33 feet to make an out at home plate. 6. a. Answers will vary.

}

5} sec B 5 } 4 adj cot B 5 } opp 5 } 5 7. h

b. Area of triangular face: 288

1

50

Area 5 }2 (6)(6) sin u 5 18 sin u V 5 Area of base p height 5 (18 sin u )(12) 5 216 sin u c. As the angle increases from 08 to 908, the volume

increases. As the angle increases from 908 to 1808, the volume decreases. d. When u 5 908, the trough has its maximum volume.

Answers will vary. Copyright © by McDougal Littell, a division of Houghton Mifflin Company.

}

5 }4 tan B 5 } adj

d 2 ø 60.52 1 952 2 2(60.5)(95) cos 139.48

7.

}

opp

60.5 sin 258 ø 0.2691 sin G 5 } 95

1

s 5 }2 (8 1 11 1 15) 5 17 }}}

Area 5 Ï17(17 2 8)(17 2 11)(17 2 15) ø 42.85

Volume 5 Area of base p height 5 42.85(0.5) ø 21.43

1 0.33ft 2 1 bag

21.43 ft3 }3 ø 65 bags The volume is about 21.43 cubic feet. Because one bag covers 0.33 cubic feet, about 65 bags of concrete are needed to cover the entire patio. Chapter 13 Review (pp. 898–900) 1. An angle is in standard position if its vertex is at the

origin, the initial side lies on the x-axis, and the angle moves counterclockwise. 2. Because 22258 and 1358 have a reference angle of

458 located in the second quadrant, they have the same terminal side and are said to be coterminal angles. 3. A unit circle has its center at the origin with a radius

of 1 unit. 4. If cos u 5 a and 0 ≤ u ≤ π, then the inverse cosine of

a equals u. 5. The law of sines is a proportion relating the ratio of the

sine of an angle and its corresponding side to the ratios of the sines of the other angles and their corresponding sides.

h

tan 288 5 } 50 h 5 50 tan 288 ø 26.59 The height of the balloon is about 26.59 meters. 29 π π radians 8. 1458 } 5 } radians 36 1808

1

2

π radians 4π 9. 2808 } 5 2} radians 9 1808

1

2

1808 4π 10. } } 5 2408 3 π radians

1

2

1808 11 π 11. } } 5 3308 6 π radians

1

2

12. Tangent is negative in Quadrant IV.

u 9 5 3608 2 3308 5 308

}

Ï3

1 Ï3

tan 3308 5 2tan 308 5 2} } 5 2} 3 13. Cosecant is negative in Quatrant IV. 24058 is coterminal

with 2458, whose terminal side lies in Quadrant IV. }

csc(24058) 5 2csc 458 5 2Ï 2

13 π π 14. Sine is positive in Quadrant I. } is coterminal with } 6 6

whose terminal side lies in Quadrant I. 13 π

π

1

5 sin 1 } 5 }2 sin 1 } 6 2 62 11 π 15. Secant is positive in Quadrant IV. } is coterminal 3 5π whose terminal side lies in Quadrant IV. with } 3 5π π u 9 5 2π 2 } 5} 3 3 11 π π sec } 5 sec } 52 3 3


n2ws13-b.indd 759

759

6/27/06 11:52:20 AM

Chapter 13,

continued

16. Sine is negative in Quadrants III and IV. π π ≤u≤} , or 2908 ≤ u ≤ 908: When 2} 2 2 π u 5 sin21(20.5) 5 2} or 2308 6 17. 12 feet

20. h 5 a sin C 5 17 sin 558 ø 13.9

Because 13.9 < 15 < 17, there are 2 triangles. Triangle 1:

B

4 feet

17

u

558

u 5 sin21 1 } 5 sin21 1 }3 2 ø 19.58 12 2 4

1

C

C 838

17 sin 558

ø 0.9284 sin A 5 } 15

a

A ø sin 21 0.9284 ø 68.28

438 A

c

B

B ø 1808 2 558 2 68.28 5 56.88

ASA: one triangle

15 b }5} sin 56.88 sin 558

B 5 1808 2 438 2 838 5 548

15 sin 56.88 sin 558

12 a }5} sin 548 sin 438

b 5 } ø 15.3 A ø 68.28, B ø 56.88, b ø 15.3

12 sin 438 sin 548 12 c }5} sin 548 sin 838

a 5 } ø 10.1

Triangle 2:

B

17

12 sin 838 sin 548

c 5 } ø 14.7

C b A

A ø 1808 2 68.28 5 111.88

C

B ø 1808 2 558 2 111.88 5 13.28

25

a

15 sin 13.28 sin 558

b 5 } ø 4.2

Because 25 > 18, there is one triangle. sin 1048 25

}5}

A ø 111.808, B ø 13.28, b ø 4.28 21.

C

18 sin 1048 ø 0.6896 sin C 5 } 25

C 5 sin210.6896 ø 44.38 A ø 1808 2 1048 2 44.38 5 31.78 a sin 31.78

25 sin 1048

}5}

25 sin 31.78 sin 1048

15 sin 558

b sin 13.28

}5}

1048 18 B

sin C 18

15

558

B 5 548, a ø 10.1, c ø 14.7 19.

A

A

20

738

a 608

A

c

B

AAS: one triangle A 5 1808 2 608 2 738 5 478 a sin 478

20 sin 608

a 5 } ø 13.5

}5}

A ø 31.78, C ø 44.38, a ø 13.5

a 5 } ø 16.9

20 sin 478 sin 608

c sin 738

20 sin 608

}5}

20 sin 738 sin 608

c 5 } ø 22.1 A ø 478, a ø 16.9, c ø 22.1

760


12

b

sin 558 sin A }5} 15 17

The angle of elevation of the ramp is about 19.58. 18.

15


n2ws13-b.indd 760

6/27/06 11:52:30 AM

Chapter 13,

continued

22. 192 5 112 1 142 2 2(11)(14) cos A

3. adj 5

44 5 2308 cos A A ø cos21(20.1429) ø 98.28

}

Ï13

sin u 5 } 7

7

6

cos u 5 }7

sec u 5 }6

sin 98.28 sin B }5} 19 11

}

}

6Ï13

6 Ï13

Ï13

tan u 5 } 6

cot u 5 } } 5 } 13

11 sin 98.28

ø 0.5730 sin B 5 } 19

13 π π radians 4. 2608 } 5 } radians 9 1808

1

B ø sin21 0.5730 ø 35.08 C ø 1808 2 98.28 2 35.08 5 46.88

2

5π π radians 5. 2508 } 5 2} radians 18 1808

1

A ø 98.28, B ø 35.08, C ø 46.88

2

1808 4π 6. } } 5 1448 5 π radians

1

23. b 2 5 202 1 172 2 2(17)(20) cos 758 ø 513.0 }

b ø Ï513.0 ø 22.6

2

8π 1808 7. } } 5 4808 3 π radians

1

sin 758 22.6

}5}

2

8. Tangent is negative in Quadrant II.

20 sin 758

ø 0.8548 sin A 5 } 22.6

u 5 1808 2 1508 5 308

A ø sin

tan 1508 5 2tan 308 5 2} } 5 2} 3

21

0.8548 ø 58.78

C ø 1808 2 758 2 58.78 5 46.38 2

2

with 2408 whose terminal side lies in Quadrant III.

2

24. 39 5 30 1 35 2 2(30)(35) cos C

u 9 5 2408 2 1808 5 608 sec(24808) 5 2sec 608 5 22

2604 5 22100 cos C 0.2876 ø cos C

10. Sine is positive in Quadrant I.

5π

C ø cos21 0.2876 ø 73.38 sin B 35

73.38 39

Quadrant I. 5π

12 π

A ø 1808 2 59.38 2 73.38 5 47.48

11 π

4 Ï97

4Ï97

sinu 5 } } 5 } 97

}

}

9 Ï97

4

}

}

}

11Ï57

csc 5 } } 5 } 57

}

Ï 57

sin u 5 } 11 8

11

cos u 5 } 11

sec u 5 } 8 }

8Ï57

cot u 5 } } 5 } 57

u 5 cos21 5 0 or 08 π π 13. When 2 } ≤ u ≤ }, or 2908 ≤ u ≤ 908: 2 2 π

}

u 5 tan21 Ï3 5 } or 608 3

π π 14. When 2} < u < }, or 2908 < u < 908: 2 2

tanu 5 }9

2. opp 5 Ï 112 2 82 5 Ï 57

8 Ï 57

9Ï97

cosu 5 } } 5 } 97

9

Ï3

0 ≤ u ≤ π, or 0 ≤ u ≤ 1808:

}

cscu 5 } 4

cotu 5 }4

π

12. Cosine is positive in Quadrants I and IV. When

}

1. hyp 5 Ï 42 1 92 5 Ï 97

Ï97

π

}

cos1 } 5 cos1 } 5} 2 6 2 62

Chapter 13 Test (p. 901)

secu 5 } 9

11 π

u9 5 } 2} 5} 6 6 6

A ø 47.48, B ø 59.38, C ø 73.38

Ï97

Ï3

11. Cosine is positive in Quadrant IV.

B ø sin21 0.8596 ø 59.38

}

}

π

sin1 2} 5 sin1 } 5} 3 2 2 32

35 sin 73.38 sin B 5 } ø 0.8596 39

}

π

2} is coterminal with } whose terminal side lies in 3 3

} 5 sin }

11 Ï 57

}

Ï3

1 Ï3

9. Secant is negative in Quadrant III. 2480 is coterminal

A ø 58.78, C ø 46.38, b ø 22.6


}

7Ï13

7 Ï13

csc u 5 } } 5 } 13

20.1429 ø cos A

sin A 20

}} }

Ï72 2 (Ï13 )2 5 6

1

}

Ï2

2

π

u 5 sin21 2} 5 2} or 2458 2 4 15. When 0 ≤ u ≤ π, or 0 ≤ u ≤ 1808:

1

}

Ï3

2

5π

u 5 cos21 2} 5} or 1508 2 6

}

Ï 57

tan u 5 } 8


n2ws13-b.indd 761

761

6/27/06 11:52:35 AM

Chapter 13,

continued

16. AAS: one triangle

20. SSS: one triangle B

B

12

a

478

328

A

b

C

12 sin 328

900 5 1017 2 1008 cos B 0.1161 ø cos B

12 sin 478 sin 328

a 5 } ø 16.6

B ø cos21 0.1161 ø 83.3

12 sin 328

} 5 } ø 0.7945

12 sin 1018 sin 328

ø0.7945 sin A 5 24 sin } 30

sin 83.38 30

sin A 24

}5}

83.38

b 5 } ø 22.2 B 5 1018, a ø 16.6, b ø 22.2

A ø sin21 0.7945 ø 52.68 C ø 1808 2 83.38 2 52.68 5 44.18

17. SSS: one triangle

A ø 52.68, B ø 83.38, C ø 44.18

B

21. ASA: one triangle

24

B

778

50 12

A

C

a

268

242 5 122 1 172 2 2(12)(17) cos A

A

576 5 433 2 408 cos A

C 5 1808 2 778 2 268 5 778

A ø cos

(20.3505) ø 110.58

}5}

50 sin 268 sin 778

a 5 } ø 22.5

17 sin 110.58

ø 0.6635 sin C 5 } 24 C ø sin

21

0.6635 ø 41.68

B ø 1808 2 110.58 2 41.68 5 27.98 A ø 110.58, B ø 27.98, C ø 41.68

C ø 778, a ø 22.5, b 5 50 1 22. Area 5 } (16)(18) sin 818 5 142.2 units2 2 23.

18. h 5 a sin B 5 11 sin 638 ø 9.8

}}} }

ø Ï413.44 ø 20.3 units2

19. SAS: one triangle C

24.

1018

23

c

1

s 5 }2 (8 1 6 1 7) 5 10.5 Area 5 Ï10.5(10.5 2 8)(10.5 2 6)(10.5 2 7)

Because 9.8 > 8 (h > b), there is no triangle.

19

50 sin 778

a sin 268

}5}

sin 110.58 24

sin C 17

C

Because this is an isosceles triangle, b 5 50.

20.3505 ø cos A 21

b

1

s 5 }2 (25 1 24 1 38) 5 43.5 }}}}

Area 5 Ï43.5(43.5 2 25)(43.5 2 24)(43.5 2 38) }

B

c 2 5 192 1 232 2 2(19)(23) cos 1018 ø 1056.77 }

c ø Ï1056.77 ø 32.5 sin 1018 sin A }5} 32.5 23 23 sin 1018 ø 0.6947 sin A 5 } 32.5

A ø sin21 0.6947 ø 44.08 B ø 1808 2 1018 2 44.08 ø 35.08 A ø 44.08, B ø 35.08, c ø 32.5

ø Ï86,309.44 ø 293.8 units2 1 25. Area 5 } (7)(13) sin 1118 ø 42.5 units2 2 26.

1

s 5 }2 (16 1 33 1 24) 5 36.5 }}}}

Area 5 Ï36.5(36.5 2 16)(36.5 2 33)(36.5 2 24) ø180.9 units2

1 27. Area 5 } (12)(18) sin 618 ø 94.5 units2 2


17

A

C

302 5 242 1 212 2 2(24)(21) cos B

}5}

b sin 1018

30

A

B 5 1808 2 478 2 328 5 1018

a sin 478

24

21

w 28. } 5 tan 398 80

w ø 64.8 The width of the river is about 64.8 meters.

762


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Chapter 13,

continued

29.

3. 3

160 ft

200 ft

u

155 ft

5

3

u

5 ft

Ground

u5

tan u 5 }5

u 5 tan21 1 }5 2 ø 31.08 3

1 200 2 ø 50.88

155 sin21 }

The angle of elevation is about 318. Because the Fixed steps has a range of angles from 308– 358, it would be the most appropriate tool for the task.

The arm should be set at an angle of about 50.88. 30.

208

40 mi

u

25 mi

d

4. 9

u 57

9 tan u 5 } 57

u 5 1808 2 208 5 1608 d 2 5 252 1 402 2 2(25)(40) cos 1608 ø 64

9

u 5 tan21 } ø9 57

The boat is about 64 miles from its point of departure.

The path of the ball must take an angle of about 98 with the ground.

Standardized Test Preparation (pp. 902–903) 1. No credit. The student did not use the correct procedure

to find the volume of the triangular prism, so the calculations, reasoning, and answers are incorrect.

5. Liquid: 151.82 5 962 1 962 2 2(96)(96) cos u

23,043.24 5 18,432 2 18,432 cos u

2. Full Credit. The student provided an accurate diagram

20.2502 ø cos u

and calculated the volume of the triangular prism correctly with an explanation of the reasoning used to solve the problem.

u ø cos21 (20.2502) ø 104.58 The obtuse angle is about 104.58. Ice: 1652 5 1012 1 1012 2 2(101)(101) cos u

Standardized Test Practice (pp. 904–905)

27,225 5 20,402 2 20,402 cos u

1 1. a. A 5 } (1.383)(1.381) sin 5.718 ø 0.095AU 2 2

u ø cos21 (20.3344) ø 109.58

93 million miles 2 0.095AU p }} 1 AU 2


20.3344 ø cos u

1

2

The obtuse angle is about 109.58. Because one angle is 104.58 and the other angle is 109.58, the angles are not the same.

ø 821.67 million square miles The area Mars swept out is about 821.67 million square miles.

6.

b. Kepler’s law states the area swept out by the path of a

planet is the same over the given time interval. Planet closer to sun

h

Planet farther from sun

5 ft

As the planet moves closer to the sun, it must move faster to sweep out the same area in the same time interval.

1

2

3608 2. a. u 5 6 rev } 5 21608 1 rev π radians

u 5 21608 1 } 5 12π 180 2 The wheel of a unicycle rotates 21608 or 12π in 6 full revolutions. b. A unicycle travels 2πr for one revolution. Because the

unicycle is traveling 6 revolutions the unicycle travels about 6(2π)(12) ø 452.4 inches.

708 30 ft

From the diagram, the height of the tree is h 1 5 feet. h

, h 5 30 tan 708 ø 82. Because the Using tan 708 5 } 30 height at eye level is 5 feet, the tree is about 82 1 5 5 87 feet tall. 402 v2 7. You: d 5 } sin 2u 5 } sin 908 5 50 32 32 v2

v2

sin 2u 5 } sin 508 5 0.024v 2 Your teammate: d 5 } 32 32 50 5 0.024v2 v ø 45.6 Your teammate needs to kick the soccer ball with an initial velocity of about 45.6 feet per second.


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Chapter 13,

continued

1 8. Because Area of a sector 5 } r 2u and arc length 5 ru : 2 10π } 5 ru 3 25π 3

1 2

25π 3

1 2

25π 3

5π 3

18.

50 feet 208 lake bottom

d a. sin 208 5 } 50

} 5 }r 2u

d 5 50 sin 208 ø 17.1

1 10π 3 2

} 5 }r }

The depth of the water is about 17.1 feet. b.

} 5 }r

50 feet

55r

2π 10π 5 5u, so u 5 } . Because r 5 5, } 3 3

u

37

cos u 5 } 50 37 u 5 cos21 1 } ø 42.38 50 2

10. B;

C ø 1808 2 858 2 278 5 688

The rope makes an angle of about 42.38 with the lake bottom.

5 sin 688

a sin 858

lake bottom

37 feet

9. C; Because the angle has its terminal side in 5π radians. Quadrant III, the angle is } 4

}5}

c. The angle the rope makes with the lake bottom will get

5 sin 858 sin 688

a 5 } ø 5.4 cm

larger as the boat travels to deeper water. 19. a.

11. B; u 9 5 3608 2 3008 5 608

s 5 ru 5π 5 xu

12. 5

b.

3

s 5 ru 14π 5 (x 1 10)u

u

5π c. 5π 5 xu → u 5 } x

}

adj 5 Ï52 2 32 5 4 4

5π

cos u 5 }5

2

14. u 5 sin21 0.5 5 308

1 15. s 5 } (5 1 7 1 10) 5 11 2 }}}

50π

14π 5 5π 1 } x 50π

}

Area 5 Ï11(11 2 5)(11 2 7)(11 2 10) 5 Ï264 ø 16.2

9π 5 } x 9xπ 5 50π

The area of the triangle is about 16.2 square units. 16.

y

50

5

5 5}9 x5} 9

(10, 24)

5π

r u

5π

9π

u5} 5} 5} 50 x 10 }

x

9

2 9π 1 50 1 50 2 9π d. A 5} } 1 10 } 2 } } } 2 9 2 9 10 10

1

21 2

1 21 2

ø 342.1 2 43.6 5 298.5 }

r 5 Ï102 1 242 5 26 26

sec u 5 } 5 2.6 10 The value of sec u is 2.6. 17.

132 5 72 1 92 2 2(7)(9) cos A 169 5 130 2 126 cos A 20.3095 ø cos A

The amount of fabric needed is about 298.5 square inches.


1

5π 1808 13. } p } 5 1508 6 π radians

14π 5 (x 1 10) } x

A ø cos21 (20.3095) ø 108.08

764


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