UK Secondary (7-11)
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Sine Rule
Lesson

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).

The sine rule

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.

 

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=sin1(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 $=$= $\sin^{-1}\left(\frac{28\times\sin39}{41}\right)$sin1(28×sin3941)
$R$R $=$= $25.5^\circ$25.5° (to 1 d.p.)

 

 

Practice questions

Question 1

Find the value of the acute angle $x$x using the Sine Rule.

Write your answer in degrees correct to one decimal place.

Question 2

Find the side length $a$a using the sine rule.

Round your answer to two decimal places.

Question 3

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.

  1. Solve for the unknown interior angle $A$A.

  2. Solve for $b$b.

    Round your answer to three decimal places.

  3. Solve for $c$c.

    Round your answer to three decimal places.

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