The sine rule and the cosine rule · Sine & cosine rules

SOH CAH TOA only works in right-angled triangles. For any other triangle there are two rules. The labelling is the same for both: side $a$ is opposite angle $A$ , side $b$ opposite $B$ , and side $c$ opposite $C$ .

Sides and opposite anglesSide a faces angle A, and so on

The sine ruleDrag the corners

12.26a ÷ sin A12.26b ÷ sin B12.26c ÷ sin C

Each side divided by the sine of the angle opposite it gives the same number. Use it when you know a side and the angle opposite it: a ÷ sin A = b ÷ sin B = c ÷ sin C.

The sine rule

Worked example · JAMB 1995

JAMB 1995 · UME · Q36

$QRS$ is a triangle with $QS = 12$ m, $\angle RQS = 30^\circ$ and $\angle QRS = 45^\circ$ . Calculate the length of $RS$ .

Match sides with opposite angles
$RS$ is opposite $\angle RQS = 30^\circ$ . $QS = 12$ is opposite $\angle QRS = 45^\circ$ . That’s a side with its opposite angle known: the sine rule.
Think first. Which side is opposite the 45° angle, and which is opposite the 30° angle?
Write the sine rule
$\frac{RS}{\sin 30^\circ} = \frac{12}{\sin 45^\circ}$
Solve with exact values
$\begin{aligned} RS &= \frac{12 \sin 30^\circ}{\sin 45^\circ} = \frac{12 \times \frac12}{\frac{\sqrt2}{2}} \\ &= \frac{12}{\sqrt2} = 6\sqrt2\text{ m} \end{aligned}$
The answer is C.

More: the sine rule

The cosine rule

Look at the board in cosine mode: when $C$ is a right angle, $\cos C = 0$ and the rule is just Pythagoras. The extra term corrects for the angle not being $90^\circ$ .

More: the cosine rule

Which rule?

You know	You want	Use
two angles and a side	another side	sine rule
two sides and a non-included angle	an angle	sine rule
two sides and the angle between them	the third side	cosine rule
all three sides	an angle	cosine rule

The angle bisector

The line that cuts an angle of a triangle in half divides the opposite side in the same ratio as the two sides next to the angle. In triangle $MNO$ , if the bisector of angle $M$ meets $NO$ at $P$ , then

\frac{NP}{PO} = \frac{MN}{MO}

The angle bisectorThe longer side next to M gets the longer piece

More: the angle bisector

Your turn

JAMB 1988 · UME · Q42

Sine & cosine rules

In triangle $PQR$ , $PQ = 1$ cm, $QR = 2$ cm and $\angle PQR = 120^\circ$ . Find the longest side of the triangle.

Worked solution (try it first)

The longest side faces the largest angle, the $120^\circ$ at $Q$ .
So the longest side is $PR$ .
Cosine rule: $PR^2 = 1^2 + 2^2 - 2(1)(2)\cos120^\circ$ .
$\cos120^\circ = -\frac12$ , so $PR^2 = 5 + 2 = 7$ and $PR = \sqrt7$ cm, option D.

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