The sine rule

In a nutshell

The sine rule is used to work out the missing lengths or angles in any type of triangle. There are two forms of the sine rule you will need to know.

The sine rule formula

The sine rule is used when you have a side-angle pair (e.g. side $b$ and angle $B$ ), and either another side or another angle. There are two forms of the sine rule: The side form and the angle form.

Side form	Angle form
$\dfrac{a}{\sin(A)}=\dfrac{b}{\sin(B)}=\dfrac{c}{\sin(C)}$	$\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}$

This is for a triangle with sides $a,b,c$ and corresponding angles $A,B,C$ .

Maths; Trigonometry; KS5 Year 12; The sine rule

Example 1

A triangle has angles $A$ and $B$ which are $40^\circ$ and $70^\circ$ respectively. If the side $a$ is $5cm$ long, what is the length of side $b$ to $1$ decimal place?

Use the sine rule to form an equation involving the given values of $A$ , $B$ , $a$ and the unknown value of $b$ :

$\dfrac{a}{\sin(A)}=\dfrac{b}{\sin(B)}$

Substitute the known values:

$\dfrac{5}{\sin(40)}=\dfrac{b}{\sin(70)}$

Solve for $b$ :

b = \sin(70)\times \dfrac{5}{\sin(40)} = 7.309511...

$\underline{b =7.3 \ cm \ (1d.p.)}$

Example 2

In $\triangle ABC$ , $AB = 12 \ cm$ , $BC=15 \ cm$ and $\angle CAB = 75 ^{\circ}$ . Find angle $B$ and $C$ .

Maths; Trigonometry; KS5 Year 12; The sine rule

To solve angles use the angle form of the sine rule:

$\dfrac{\sin(A)}{a}=\dfrac{\sin(B)}{b}=\dfrac{\sin(C)}{c}$

Substitute the known values:

$\dfrac{\sin(75)}{15} = \dfrac{\sin(B)}{12}$

Solve for angle $B$ :

$\dfrac{12\sin(75)}{15}= \sin(B)$

$B= \sin^{-1}(\dfrac{12\sin(75)}{15}) =50.60063...$

$\underline{B=50.6^{\circ}(1d.p.)}$

Angles in a triangle add up to $180^{\circ}$ . Use this to find angle $C$ :

$180 -50.6 -75=$

$\underline{C=54.4^{ \circ} (1.dp)}$