Sine Rule and Cosine Rule
Sine and Cosine Rules

In the triangle ABC, the side opposite angle A has length a, the side opposite angle B has length b and the side opposite angle C has length c.

The sine rule states

= =

Proof of Sine Rule

If you construct the perpendicular from vertex A to meet side CB at N, then

AN = csinB (from Δ ABN)
= bsinC (from Δ ACN)

Hence

csinB = bsinC ⇒ =

similarly for .

The cosine rule states

a2 = b2 + c2 − 2bccosA

b2 = c2 + a2 − 2cacosB

c2 = a2 + b2 − 2abcosC

Proof of Cosine Rule

If CN = x, then NB = a − x and

c2 = AN2 + (a − x)2 when x = CN
= (bsinC)2 + (a − bcosC)2, since x = bcosC
= b2sin2C + b2cos2C − 2abcosC + a2
= b2(sin2C + cos2C) + a2 − 2abcosC

i.e. c2 = b2 + a2 − 2abcosC, since sin2C + cos2C = 1

Worked Examples

1

Find the unknown angles and side length of the triangle shown.

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Using the sine rule,

= =

From the first equality,

sinA = = 0.5638
A = 34.32°

Since angles in a triangle add up to 180° ,

B = 180° − 70° − A = 75.68°

From the sine rule,

=
b =
=
= 3.61 cm
2

Find two solutions for the unknown angles and side of the triangle shown.

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Using the sine rule,

= =

From the second equality,

sinB = = 0.8030

A graph of sinx shows that between 0° and 180° there are two solutions for B.

These solutions are B = 53.41° and, by symmetry,

B = 180 − 53.41 = 126.59°

Solving for angle A we have

A = 180° − 42° − B

when B = 53.41°, A = 84.59°

when B = 126.59°, A = 11.41°

From the sine rule,

a =

For A = 84.59°, B = 53.41°, a = 7.44 cm

For A = 11.41°, B = 126.59°, a = 1.48 cm
3

Find the unknown side and angles of the triangle shown.

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To find a, use the cosine rule:

a2 = 4.92 + 3.72 − 2 × 4.9 × 3.7 × cos65°

a2 = 22.3759

a = 4.73 (to 2 d.p.)

To find the angles, use the sine rule:

= =

sinB = = = 0.9389
B = 69.86°
sinC = = = 0.7090
C = 45.15° (alternatively, use A + B + C = 180° to find C)

Checking, A + B + C = 65°+ 69.86°+ 45.15°= 180.01°. The three angles should add to 180°; the extra 0.01° is due to rounding errors.
4

The diagram below, not drawn to scale, shows the journey of a ship which sailed from Port A to Port B and then to Port C. Port B is located 32 km due West of Port A and Port C is 45 km from Port B on a bearing of 040° .

Calculate, giving your answers correct to 3 significant figures

(a)

the distance AC

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Using the COSINE rule,

AC2 = BC2 + BA2 − 2 × BC × BA × cosCBA

As angle CBA = 90° − 40°, then

AC2 = 322 + 452 − 2 × 32 × 45 × cos50°
≈ 1197.77
AC ≈ 34.6088
= 34.6 to 3 significant figures
(b)

the bearing of Port C from Port A.

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The bearing of C from A is

270° + angle BAC

Using the SINE rule, =

sinBAC =
≈ 0.9960
So angle BAC ≈ 84.903°
= 84.9° to 3 significant figures

Hence the bearing of Port A from Port B is

270° + 84.9° = 354.9°
= 355° to 3 significant figures

Some important values of sinθ, cosθ and tanθ are shown in this table.

θsinθcosθtanθ
0°010
30°
45°1
60°
90°10infinite

The graphs of sinθ and cosθ for any angle are shown in the following diagrams.

The graphs are examples of periodic functions. Each basic pattern repeats itself every 360°. We say that the period is 360°.

5

An oil tanker leaves Town X, and travels on a bearing of 050° to Town Z, 50 km away. The tanker then travels to Town Y, 70 km away, on a bearing of 120° .

(a)

Draw a carefully labelled diagram of the tanker's journey, clearly showing the North line.

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(b)

Calculate the distance of Y from X, giving your answer to 3 significant figures.

Show me In triangle XZY, angle XZY = 50° + 60°= 110° so, using the cosine rule,
XY2 = 502 + 702 − 2 × 50 × 70 × cos110°
= 2500 + 4900 − 7000cos110°
≈ 7400 − 7000 × (−0.3420)
≈ 9794.14
XY ≈ 98.965
= 99.0 to 3 significant figures.
(c)
(i)

On your diagram, mark the angle that shows the bearing of X from Y.

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As marked on diagram.
(ii)

Calculate the bearing of X from Y, giving your answer to the nearest degree.

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We need to find the angle ZYX to determine the bearing of X from Y.

In triangle ZYX, using the sine rule,

≈

sinZYX ≈ 0.4748

angle ZYX ≈ 28.34° = 28° to the nearest degree

Hence the bearing of X from Y

= 360° − (60° + 28°)

= 272°
6

Find the shaded angle in the triangle shown.

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Using the cosine rule,

252 = 162 + 132 − 2 × 16 × 13cosx

Rearranging,

2 × 16 × 13cosx = 162 + 132 − 252
416cosx = −200
cosx =
x ≈ cos−1⎛
⎝⎞
⎠°
= 118.7° to 1 decimal place
using the SHIFT and COS buttons on a calculator.

Note

As cosx is negative, the angle will be obtuse (90° ≤ x ≤ 180°).

Exercises

For each of the triangles, find the unknown angle marked θ.

Round your answers to 1 d.p.

(a)
θ = °
(b)
θ = °
(c)
θ = °
(d)
θ = °
(e)
θ = °
(f)
θ = °

For each triangle, find the unknown side marked a, b or c.

Round to 2 d.p.

(a)
a =
(b)
b =
(c)
a =
(d)
c =
(e)
a =
(f)
a =

For each of the triangles, find the unknown angles and sides.

Round degrees to 1 d.p. and lengths to 2 d.p.

(a)

A = °

B = °

a =

(b)

A = °

B = °

b =

(c)

B = °

b =

c =

(d)

A = °

B = °

a = cm

Find the remaining angles and sides of the triangle ABC if B = 81°, b = 12 and c = 11.

A = ° (to 1 d.p.)

C = ° (to 1 d.p.)

a = (to 2 d.p.)

For each of the following triangles, find the unknown angles and sides.

Round degrees to 1 d.p. and lengths to 2 d.p.

(a)

B = °

C = °

a =

(b)

B = °

C = °

a =

(c)

A = °

B = °

C = °

(d)

A = °

B = °

C = °

(e)

A = °

C = °

b =

(f)

A = °

B = °

c =

To calculate the height of a church tower, a surveyor measures the angle of elevation of the top of the tower from two points 50 metres apart. The angles are shown in the diagram.

Round to 1 d.p.

(a)

Calculate the distance BC.

m
(b)

Hence calculate the height of the tower CD.

m

The diagram shows a crane working on a wharf. AB is vertical.

Round to 1 d.p.

(a)

Find the size of angle ABC.

°
(b)

Find the height of point C above the wharf.

m

The rectangular box shown in the diagram has dimensions 10 cm by 8 cm by 6 cm. Find the angle θ formed by a diagonal of the base and a diagonal of the 8 cm by 6 cm side.

° (to 2 d.p.)

In triangle ABC, AC = 12.6 cm, BC = 11.2 cm and angle B = 54°. The lengths AC and BC are correct to the nearest millimetre and angle B is correct to the nearest degree. Use the sine rule

=

to calculate the smallest possible value of angle A.

° (to 2 d.p.)

The banks of a river are straight and parallel. To find the width of the river, two points, A and B, are chosen 50 metres apart. The angles made with a tree at C on the opposite bank are measured as angle CAB = 56°, angle CBA = 40°. Calculate the width of the river.

m (to 1 d.p.)