If we have a rightangled 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.
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 rightangled 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).
For a triangle with sides $a$a, $b$b, and $c$c, with corresponding angles $A$A, $B$B, and $C$C,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$sinAa=sinBb=sinCc.
We can also take the reciprocal of each fraction to give the alternate form,
$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$asinA=bsinB=csinC.
The sine rule shows that the lengths of the sides in a triangle are proportional to the sines of the angles opposite them.
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.
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.)  

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).
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.)  

Find the value of the acute angle $x$x using the Sine Rule.
Write your answer in degrees correct to one decimal place.
Find the side length $a$a using the sine rule.
Round your answer to two decimal places.
Consider the triangle with two interior angles $C=72.53^\circ$C=72.53° and $B=31.69^\circ$B=31.69°, and one side length $a=5.816$a=5.816 metres.
Solve for the unknown interior angle $A$A.
Solve for $b$b.
Round your answer to three decimal places.
Solve for $c$c.
Round your answer to three decimal places.
Apply trigonometric relationships, including the sine and cosine rules, in two and three dimensions
Apply trigonometric relationships in solving problems