Sine rule

If we have a right-angled triangle, we can use trigonometric ratios to relate the sides and angles:

Here, $\sin A=\frac{a}{c}$sinA=ac​ and $\sin B=\frac{b}{c}$sinB=bc​.

But what happens when we have a different kind of triangle?

In a triangle like this, the same equations do not hold. We need to think of a different way to relate the sides and angles together.

 

The sine rule

Let's start by drawing a line segment from the vertex $C$C perpendicular to the edge $c$c. We'll call the length of this segment $x$x.

Since $x$x is perpendicular to $c$c, the two line segments meet at right angles. This means that we have divided our triangle into two right-angled triangles, and we can use the equations we already know. The relationships for the sines of the angles $A$A and $B$B is given by

$\sin A=\frac{x}{b}$sinA=xb​ and $\sin B=\frac{x}{a}$sinB=xa​.

However, $x$x wasn't in our original triangle. So we want to find a relationship using only $A$A, $B$B, $a$a and $b$b. Multiplying the first equation by $b$b and the second by $a$a gives us

$x=b\sin A$x=bsinA and $x=a\sin B$x=asinB,

and equating these two equations eliminates the $x$x and leaves us with

$b\sin A=a\sin B$bsinA=asinB.

Dividing this last equation by the side lengths gives us the relationship we want:

$\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa​=sinBb​.

We can repeat this process to find how these two angles relate to $c$c and $C$C, and this gives us the sine rule (sometimes called the law of sines).

 

Finding a side length using the sine rule

Suppose we had the angles $A$A and $B$B and the length $b$b and we wanted to find the length $a$a. Using the form of the sine rule with numerator lengths $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA​=bsinB​, we can make $a$a the subject by multiplying both sides by $\sin A$sinA. This gives

$a=\frac{b\sin A}{\sin B}$a=bsinAsinB​.

 

Worked example

Solve: Find the length of $PQ$PQ to two decimal places.

Think: The side we want to find is opposite a known angle, and we also know a matching side and angle. This means we can use the sine rule.

Do:

$\frac{PQ}{\sin48^\circ}$PQsin48°​	$=$=	$\frac{18.3}{\sin27^\circ}$18.3sin27°​
$PQ$PQ	$=$=	$\frac{18.3\sin48^\circ}{\sin27^\circ}$18.3sin48°sin27°​
$PQ$PQ	$=$=	$29.96$29.96 (to 2 d.p.)

 

Finding an angle using the sine rule

Suppose we had the side lengths $a$a and $b$b, and the angle $B$B, and we want to find the angle $A$A. Using the form of the sine rule with numerator sines $\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa​=sinBb​, we first multiply both sides by $a$a. This gives $\sin A=\frac{a\sin B}{b}$sinA=asinBb​. We then take the inverse sine of both sides to make $A$A the subject, which gives

$A=\sin^{-1}\left(\frac{a\sin B}{b}\right)$A=sin−1(asinBb​).

Worked example

Solve: Find $\angle PRQ$∠PRQ to one decimal place.

Think: The angle we want to find is opposite a known side, and we also know a matching angle and side. This means we can use the sine rule.

Do:

$\frac{\sin R}{28}$sinR28​	$=$=	$\frac{\sin39^\circ}{41}$sin39°41​
$\sin R$sinR	$=$=	$\frac{28\times\sin39^\circ}{41}$28×sin39°41​
$R$R	$=$=	$\sin^{-1}\left(\frac{28\times\sin39}{41}\right)$sin−1(28×sin3941​)
$R$R	$=$=	$25.5^\circ$25.5° (to 1 d.p.)

 

Practice questions

Question 1

Question 2

Question 3