I.G.C.S.E. Sine and Cosine Rules Index: Please click on the ...

I.G.C.S.E. Sine and Cosine Rules Index: Please click on the question number you want Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

You can access the solutions from the end of each question

Question 1 For each of the following find each side marked with the letter. a.

C

b A

B

b. a

35

!

c

6 cm 23!

B

7.1 cm C

56!

5.6 cm

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A

Solution to question 1 a.

a b c = = sin A sin B sin C

C

b

a

35!

A

6 cm 23!

c

B

Using

b a we have = sin B sin A

b 6 6 sin23! = ⇒ b = = 4.09 cm sin35! sin 23! sin35!

C = 180! − 35! − 23! = 122! c a c 6 6 sin122! = ⇒ = ⇒ c = = 8.87 cm sin C sin A sin122! sin35! sin35!

b.

B

a

c 7.1 cm

C

b

Here we have two sides and the included angle so we have to use the cosine rule. a 2 = b 2 + c 2 − 2bc cos A = 5.62 + 7.12 − 2 ( 5.6 )( 7.1) cos56!

56!

= 37.302…

5.6 cm A

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a = 37.302… = 6.11cm

Question 2 In each of the following triangles find the missing angles C

a.

b.

C 4.2 cm

B

4.5 cm

9.8 cm 84!

B A

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7.6 cm

5.7 cm

A

Solution to question 2 a.

C

a

b

4.2 cm

9.8 cm 84!

B

c

A

sin A sin B sin C = = a b c sin A sin B we have Using = a b sin A sin84! 4.2 sin84! = ⇒ sin A = 4.2 9.8 9.8 !  4.2 sin84  ! A = sin−1   = 25.2 9.8  

C = 180! − 84! − 25.2! = 70.8!

b.

Here we have three sides and we are required to find an angle, so we use the cosine rule.

C a 4.5 cm

b B 5.7 cm

b2 + c 2 − a2 2bc 2 7.6 + 5.72 − 4.52 = 2 ( 7.6 )( 5.7 )

cos A =

7.6 cm

A

c

 7.62 + 5.72 − 4.52  A = cos −1    2 ( 7.6 )( 5.7 )  = 36.1!

a2 + c 2 − b2 2ac 2 4.5 + 5.72 − 7.62 = 2 ( 4.5 )( 5.7 )

cos B =

 4.52 + 5.72 − 7.62  B = cos −1    2 ( 4.5 )( 5.7 )  = 95.6! C = 180! − 36.1 ! − 95.6! = 48.3!

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Question 3 A destroyer D and a frigate F leave a port P at the same time. The diagram shows the destroyer, sailing 45 km on a bearing of 050! and the frigate sailing 35 km on a bearing of 165! . How far are the ships apart? N D

N 45 km W

50!

E

S

P

165!

35 km

F

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Solution to question 3 N D

N 45 km W

E

50!

P

S

165!

35 km

F D f

We have to calculate DF. Drawing triangle PDF, we have Pˆ = 165! − 50! = 115! . Using the cosine rule as we have two sides and the included angle.

45 km

P

115!

p

p 2 = d 2 + f 2 − 2df cos P = 352 + 452 − 2 ( 35 )( 45 ) cos115!

35 km

= 4581.247…

d F

p = 4581.247… = 67.7km

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Question 4

Find all the angles of a triangle whose sides are in the ratio 4 : 5 : 6. Click here to read the solution to this question Click here to return to the index

Solution to question 4

First draw a diagram and label the sides. A

b

5

4

C

6

c

B a

Now calculating angle Aˆ and Bˆ using the cosine rule, we have b2 + c 2 − a2 2bc 2 4 + 52 − 62 = 2 ( 4 )( 5 )

cos Aˆ =

a2 + c 2 − b2 cos Bˆ = 2ac 2 6 + 52 − 4 2 = 2 ( 6 )( 5 )

5 40 1 = 8 =

 1 Aˆ = cos −1   8 ! = 82.8

45 60 3 = 4 =

3 Aˆ = cos −1   4 = 41.4!

Now using that the angle sum of a triangle is 180! , we have Cˆ = 180! − 82.8! − 41.4! = 55.8! Click here to read the question again Click here to return to the index

Question 5

From a lighthouse L an oil tanker A is 12 km away on a bearing of 126! and a car ferry C is 15 km away on a bearing of 210! . Draw a diagram clearly showing L, A and C. Find a.

the distance between A and C

b.

the bearing of A from C

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Solution to question 5

First drawing a diagram N

126!

L

12 km

210!

15 km A C

Now drawing triangle LAC. Note that Lˆ = 210! − 126! = 84! . L a

84!

c 12 km

15 km A l

C

a.

The distance between the two ships is given by AC = l . Using the cosine rule l 2 = a 2 + c 2 − 2ac cos L = 152 + 122 − 2 (15 )(12 ) cos84! = 331.369… l = 331.369… = 18.2km

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b.

N

L

126!

θ

N

12 km

210!

15 km

α

A

β

C

The required bearing is α + β as shown in the above diagram. θ = 360! − 210! = 150! ⇒ α = 180! − 150! = 30! (allied angles). Considering triangle CLA in part a L a

c 12 km

84!

15 km A

β l

C

18.203 . . . km

a2 + l 2 − c 2 2al 2 15 + 18.205…2 − 122 = 2 (15 )(18.203…)

Using the cosine rule cos β = cos C =

152 + 18.205…2 − 122  C = cos−1    2 (15 )(18.203…)  = 41.0! The required angle is α + β = 30! + 41! = 71! The bearing of A from C is 071! . Click here to read the question again Click here to return to the index

Question 6

An airliner flies from Heathwick airport H, 400 km on a bearing of 135! to A and then 500 km on a bearing of 260! to B, and then returns to the airport at H. a.

Draw a diagram, showing clearly H, A and B.

b.

Calculate the length and bearing of the return journey from B to H.

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Solution to question 6 a.

First draw a diagram. N

135!

H

N

400 km

A 260!

B

500 km

b.

N

H

135!

N

400 km b

a

α β

B

h

A

260!

500 km

From the diagram α = 180! − 135! = 45! therefore β = 360! − 45! − 260! = 55! We are required to find BH = a Using the cosine rule we have a 2 = b 2 + h 2 − 2bh cos A = 4002 + 5002 − 2 ( 400 )( 500 ) cos55! = 180569.4255 a = 180569.4255 = 425km Click here to continue with the question or go to next page

135!

θ H

400 km b

a

α B

N

55!

h

A

260!

500 km

a2 + b2 − h2 2ab 424.93…2 + 4002 − 5002 = 2 ( 424.93…)( 400 )

Using cosine rule cos H =

 424.93…2 + 4002 − 5002  H = cos −1    2 ( 424.93…)( 400 )  = 74.5! The bearing of H from B is given by α Now θ = 360! − 74.5! − 135! = 150.5! and hence α = 180! − 150.5! = 29.5! The bearing of H from B is 029.5! Click here to read the question again Click here to return to the index