Sine and cosine rules: Grade 12 Mathematics

By the end of this lesson you should be able to

Choose between the sine rule and the cosine rule for a given triangle
Find unknown sides and angles in any triangle
Find the area of a triangle using two sides and the included angle
Apply the rules to bearing problems

Why we need these rules

The ratios sine, cosine and tangent from right-angled triangles only work when the triangle has a right angle. Many real problems, such as distances across a river or the position of a ship, involve triangles with no right angle. The sine rule and the cosine rule solve those triangles.

Labelling the triangle

Label the angles with capital letters $A$ , $B$ , $C$ , and label each side with the small letter of the angle opposite it. Side $a$ is opposite angle $A$ , side $b$ is opposite angle $B$ , and side $c$ is opposite angle $C$ .

Each small letter is opposite the capital letter with the same name.

Key ideas

Sine rule

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Cosine rule

a^2 = b^2 + c^2 - 2bc\cos A \qquad\text{or}\qquad \cos A = \frac{b^2 + c^2 - a^2}{2bc}

Area of a triangle

\text{Area} = \tfrac{1}{2}ab\sin C

Use this table to decide which rule to use.

You are given	Use
Two angles and any side	Sine rule
Two sides and an angle not between them	Sine rule
Two sides and the angle between them	Cosine rule
All three sides	Cosine rule

Only two of the three fractions in the sine rule are used at a time. Choose the two that contain the information you have and the value you want.

Worked examples

Example 1: sine rule, finding a side. In triangle $ABC$ , $A = 40°$ , $B = 65°$ and $a = 8$ cm. Find $b$ .

We know a side and its opposite angle ( $a$ and $A$ ), and we want $b$ , whose opposite angle $B$ is known.

\frac{b}{\sin 65°} = \frac{8}{\sin 40°} \quad\Rightarrow\quad b = \frac{8 \sin 65°}{\sin 40°} = 11.3 \text{ cm}

Example 2: sine rule, finding an angle. In triangle $ABC$ , $a = 12$ cm, $b = 9$ cm and $A = 70°$ . Find angle $B$ .

\frac{\sin B}{9} = \frac{\sin 70°}{12} \quad\Rightarrow\quad \sin B = \frac{9 \sin 70°}{12} = 0.7048

B = \sin^{-1}(0.7048) = 44.8°

Side $b$ is shorter than side $a$ , so angle $B$ is smaller than angle $A$ . That means $B$ must be acute, so this is the only answer.

Example 3: cosine rule, finding a side. In triangle $ABC$ , $b = 7$ cm, $c = 10$ cm and $A = 60°$ . Find $a$ .

The known angle is between the two known sides, so use the cosine rule:

a^2 = 7^2 + 10^2 - 2(7)(10)\cos 60° = 49 + 100 - 70 = 79

a = \sqrt{79} = 8.89 \text{ cm}

Example 4: cosine rule, finding an angle. A triangle has sides $5$ cm, $6$ cm and $7$ cm. Find its largest angle.

The largest angle is opposite the longest side. Call it $C$ , with $c = 7$ :

\cos C = \frac{5^2 + 6^2 - 7^2}{2(5)(6)} = \frac{12}{60} = 0.2 \quad\Rightarrow\quad C = 78.5°

Example 5: area. Two sides of a triangle are $8$ cm and $11$ cm, and the angle between them is $42°$ . Find the area.

\text{Area} = \tfrac{1}{2}(8)(11)\sin 42° = 29.4 \text{ cm}^2

Example 6: bearings. A ship sails 12 km on a bearing of $040°$ , then 9 km on a bearing of $110°$ . How far is it from its starting point?

Between the two legs the ship turns through $110° - 40° = 70°$ . The angle inside the triangle at the turning point is therefore $180° - 70° = 110°$ .

The distance $d$ is opposite that angle, and the two known sides enclose it, so use the cosine rule:

d^2 = 12^2 + 9^2 - 2(12)(9)\cos 110° = 225 + 73.88 = 298.88

d = \sqrt{298.88} = 17.3 \text{ km}

Common mistakes

Calculator in the wrong mode. Make sure your calculator is in degrees. A quick test: $\sin 30°$ should give $0.5$ .

Rounding too early. Keep full calculator values through a calculation and round only the final answer.

Using the wrong rule. If you are given two sides and the angle between them, the sine rule cannot start, because no side and its opposite angle are both known. Use the cosine rule.

Forgetting that $\cos$ can be negative. For an obtuse angle such as $110°$ , $\cos 110°$ is negative, so $-2bc\cos A$ becomes positive. Let the calculator handle the sign.

Practice questions

Give lengths and areas to 3 significant figures and angles to 1 decimal place.

In triangle $PQR$ , $P = 50°$ , $Q = 70°$ and $p = 9$ cm. Find $q$ .
In triangle $ABC$ , $a = 15$ cm, $b = 10$ cm and $C = 48°$ . Find $c$ .
A triangle has sides of $8$ cm, $10$ cm and $12$ cm. Find its smallest angle.
Find the area of a triangle with sides of $7$ cm and $9$ cm and an included angle of $30°$ .
In triangle $ABC$ , $a = 14$ cm, $b = 10$ cm and $A = 40°$ . Find angle $B$ .
A boat sails 8 km on a bearing of $060°$ , then 5 km on a bearing of $140°$ . Find its distance from the start.
Two sides of a triangle are $6$ cm and $8$ cm and its area is $20$ cm². Find the acute angle between the two sides.

Show answers

$q = \dfrac{9 \sin 70°}{\sin 50°} = 11.0$ cm
$c^2 = 15^2 + 10^2 - 2(15)(10)\cos 48° = 124.26$ , so $c = 11.1$ cm
The smallest angle is opposite the $8$ cm side: $\cos\theta = \dfrac{10^2 + 12^2 - 8^2}{2(10)(12)} = 0.75$ , so $\theta = 41.4°$
$\tfrac{1}{2}(7)(9)\sin 30° = 15.75$ cm², which is $15.8$ cm² to 3 significant figures
$\sin B = \dfrac{10 \sin 40°}{14} = 0.4591$ , so $B = 27.3°$ . Side $b$ is shorter than side $a$ , so $B$ is acute.
The turn is $140° - 60° = 80°$ , so the interior angle is $100°$ . Then $d^2 = 8^2 + 5^2 - 2(8)(5)\cos 100° = 102.89$ , so $d = 10.1$ km
$\tfrac{1}{2}(6)(8)\sin\theta = 20$ , so $\sin\theta = \dfrac{20}{24} = 0.8333$ and $\theta = 56.4°$

Summary

Label each side with the small letter of the angle opposite it.
Use the sine rule when you know a side and its opposite angle. Use the cosine rule for two sides and the included angle, or for three sides.
Area $= \tfrac{1}{2}ab\sin C$ when you know two sides and the angle between them.
In bearing problems, work out the angle inside the triangle first, then choose the rule.

Now practise this topic

Real questions show you what still needs work. Try a past paper under timed conditions, or take a short quiz.

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