5.1 The sine and cosine rules and the area of a triangle — Learn

Prerequisites: right-angled trigonometry with sine, cosine and tangent; Pythagoras' theorem; rearranging a formula; using a calculator in degree mode.

By the end of this section you can: use the sine rule to find a side or an angle; use the cosine rule to find a side or an angle; find the area of any triangle from two sides and the included angle; decide which rule a question needs.

1. Labelling a triangle

The rules in this section work for every triangle, not only right-angled ones. Label the sides with lower-case letters and the angles with the matching capital letters. Side $a$ is opposite angle $A$ , side $b$ is opposite angle $B$ , and side $c$ is opposite angle $C$ .

The names matter. The sine rule pairs each side with the angle opposite it.

2. The sine rule

For any triangle,

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

Use the sine rule when you know a matching side and angle, and one more piece of the triangle. There are two shapes of question:

two angles and any side, to find another side;
two sides and an angle opposite one of them, to find another angle.

Turn the rule upside down when you are finding an angle:

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

Example 1. In triangle $ABC$ , angle $A = 40^\circ$ , angle $B = 65^\circ$ and side $a = 8$ cm. Find side $b$ .

The angle opposite $b$ is $B$ , so pair $a$ with $A$ and $b$ with $B$ :

\frac{b}{\sin 65^\circ} = \frac{8}{\sin 40^\circ}.

Multiply both sides by $\sin 65^\circ$ :

b = \frac{8\sin 65^\circ}{\sin 40^\circ} \approx \frac{8 \times 0.9063}{0.6428} \approx 11.3 \text{ cm}.

The side opposite the larger angle is longer, which agrees with the answer.

Example 2. In triangle $ABC$ , side $a = 7$ cm, side $b = 9$ cm and angle $A = 35^\circ$ . Find angle $B$ .

Use the angle form of the rule:

\frac{\sin B}{9} = \frac{\sin 35^\circ}{7}.

Multiply by $9$ :

\sin B = \frac{9\sin 35^\circ}{7} \approx \frac{9 \times 0.5736}{7} \approx 0.7375.

So $B \approx \sin^{-1}(0.7375) \approx 47.5^\circ$ .

3. The cosine rule

For any triangle,

a^2 = b^2 + c^2 - 2bc\cos A.

Use the cosine rule when you cannot use the sine rule:

two sides and the angle between them, to find the third side;
all three sides, to find an angle.

Rearrange the formula to make the angle the subject:

\cos A = \frac{b^2 + c^2 - a^2}{2bc}.

Example 3. In triangle $ABC$ , side $b = 6$ cm, side $c = 9$ cm and the included angle $A = 55^\circ$ . Find side $a$ .

Substitute the three known values:

a^2 = 6^2 + 9^2 - 2 \times 6 \times 9 \times \cos 55^\circ.

Work out each term:

a^2 = 36 + 81 - 108 \times 0.5736 \approx 117 - 61.9 = 55.1.

So $a \approx \sqrt{55.1} \approx 7.4$ cm.

Example 4. A triangle has sides $a = 8$ cm, $b = 10$ cm and $c = 12$ cm. Find angle $A$ .

Rearrange the cosine rule with $a$ on its own:

\cos A = \frac{10^2 + 12^2 - 8^2}{2 \times 10 \times 12} = \frac{100 + 144 - 64}{240} = \frac{180}{240} = 0.75.

So $A = \cos^{-1}(0.75) \approx 41.4^\circ$ .

4. The area of a triangle

The area is half the product of two sides and the sine of the angle between them:

\text{area} = \tfrac{1}{2}ab\sin C.

The angle must be the one inside the two sides you choose.

Example 5. Find the area of the triangle with $a = 7$ cm, $b = 10$ cm and included angle $C = 50^\circ$ .

\text{area} = \tfrac{1}{2} \times 7 \times 10 \times \sin 50^\circ = 35 \times 0.7660 \approx 26.8 \text{ cm}^2.

5. Try it: does the information fit one triangle or two?

Predict first. Sometimes two different triangles share the same two sides and the same non-included angle. In a triangle you know angle $A$ , side $a$ and side $b$ . For which values of $a$ do two triangles fit these measurements, and for which values does none fit? Write your prediction down before you touch the sliders.

Test. Move the sliders for angle $A$ and side $a$ . The circle has radius $a$ and centre $C$ , so each place it meets the ray from $A$ is a possible position of $B$ . The panel reports the height and how many triangles fit.

Explain. Find a setting where the circle meets the ray twice. What do the two angles at $B$ add up to, and why can the calculator give only one of them? Write two sentences.

Progress check

Return to the chapter problem. The survey gives $SL = 10$ km, with angle $30^\circ$ at $S$ and angle $105^\circ$ at $L$ .

Work out the angle at $B$ .
Use the sine rule to find the side $LB$ , opposite the $30^\circ$ angle.
Use the sine rule to find the side $SB$ , opposite the $105^\circ$ angle.
Find the area of triangle $SLB$ .

The exact value of $\sin 105^\circ$ is not yet on your list of special angles. Section 5.8 will prove that $\sin 105^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}$ . For now, use your calculator and give each answer to three significant figures.

Watch out. Three mistakes are common with these rules.

Pairing the wrong side and angle. In the sine rule every side is paired with the angle opposite it. Writing $\dfrac{a}{\sin B}$ uses the wrong angle. Draw the triangle and mark which side faces which angle before you substitute.

Choosing the sine rule when the cosine rule is needed. If the angle you know sits between the two sides you know, use the cosine rule. The sine rule cannot start without a matching side and angle.

The ambiguous case. When you use the sine rule to find an angle, two angles between $0^\circ$ and $180^\circ$ can share the same sine. The calculator gives only the acute one. Check whether the obtuse answer $180^\circ - B$ also fits the triangle. If the angles would then add to more than $180^\circ$ , it does not fit. Re-check: after finding an angle, add all three angles and confirm that they total $180^\circ$ .

You should now be able to

label a triangle so that each side is opposite its matching angle;
use the sine rule to find a side or an angle;
use the cosine rule to find a side or an angle;
decide between the two rules from the information given;
find the area of a triangle from two sides and the included angle;
check an angle answer by adding the three angles to $180^\circ$ .