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© T Madas

The Cosine Rule

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A B

C

ab

c

a2 = b2 + c2 – 2bccosAb2 = a2 + c2 – 2accosB

c2 = a2 + b2– 2abcosCThe cosine rule

in any triangle

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R P

Q

rp

q

p2 = r2 + q2– 2r qcosPThe cosine rule

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x 9

45° 65°

70°

d9

8

d 2 = 92 + 82 – 2 x 8 x cos45°

The cosine rule

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x 4

41° 68°

71°

x 4

7

x 2 = 42 + 72 – 2 x 7 x cos68°

The cosine rule

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45° 72°

63°

x

6 cm

8 cm

Calculate the missing length in the triangle below:

UsingTheCosineRule

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Calculate the missing length in the triangle below:

45° 72°

63°

x

6 cm

8 cmx 6x

2 = 62 + 82 – 2 x 8 x cos72° Û

x 2 = 36 + 64 – 96cos72° Û

x 2 = 100 – 96cos72° Û

x = Û- °100 96 cos 72

x ≈ 8.4 cm

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Calculate the missing angle in the triangle below:

θ

5 cm 6 cm

8 cm

Û=26

48.5» °

25 + 28 - 2´ 5´ 8´ cos

Û=36 25 +64 - 80 cos

Û=80 cos 25 +64 - 36

Û=80 cos 53

Û=cos 5380

= ( )-1 5380cos Û

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The Sine Rule

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=

A B

ab

c

The Sine Rulesin A sinB

=sinC

=a b

=c

C

in any triangle

sin A sinB sinC

a b c

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Calculate the missing length in the triangle below:

45° 72°

63°

x

6 cmUsingTheSineRule

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°sin72 °sin45

Calculate the missing length in the triangle below:

45° 72°

63°

x

6 cm

Ûx = 6

x °sin45 = 6 °sin72 Û

x =°

°6sin72sin45

Û

8.07 cmx »

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°sin70 °sin35

Calculate the missing length in the triangle below:

35° 75°

70°

x

8 cm

Ûx = 8

x °sin35 = 8 °sin70 Û

x =°

°8sin70sin35

Û

» 13.11 cmx

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7 5

Calculate the missing angle in the triangle below:

35°

x

7 cm

5 cm

Ûsinx = °sin35

5 sinx = 7 °sin35 Û

sinx =°7sin35

5Û

» °53x

sinx » 0.803 Û

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A soldier walked from his base for 4 km on a bearing of 060° to a point A.He then walked a further 5 km due east to a point B.Find:1. The distance of point B from the base.2. The bearing of B as measured from the base.3. The bearing of the base as measured from B.

Base

N

4060°

A 5 B

C

30°

d

150°

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Base

N

060°

A 5 B

C

30°

d

150°

– 40

x 5d 2= 52+42 – 2 x 4 x

cos150°

By the cosine rule on ABC

d 2= 25 + 16 cos150°

d 2 ≈ 75.64

d ≈8.7 km

4

8.7

© T Madas

A soldier walked from his base for 4 km on a bearing of 060° to a point A.He then walked a further 5 km due east to a point B.Find:1. The distance of point B from the base.2. The bearing of B as measured from the base.3. The bearing of the base as measured from B.

Base

N

4060°

A 5 B

C

30° 150°

8.7

8.7 km

θ

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Base

N

4060°

A 5 B

C

30° 150°

8.7θ

By the sine rule on ABC :

sinθ5

sin150°8.7

=

5sin150°8.7

sinθ =

x 55 x

0.287sinθ ≈

sin-1 (0.287)θ ≈

17°θ ≈

17°

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Base

N

4060°

A 5

C

30° 150°

8.7

A soldier walked from his base for 4 km on a bearing of 060° to a point A.He then walked a further 5 km due east to a point B.Find:1. The distance of point B from the base.2. The bearing of B as measured from the base.3. The bearing of the base as measured from B.

8.7 km077°

17°

B is at a bearing of 077° from the base

257°

B

77°

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