 # 7.05 The sine rule

Lesson

So far we have explored the relationship between angles and sides in right-angled triangles. As long as we have a right-angled triangle, Pythagoras' theorem and trigonometric ratios can help us to find missing side lengths, and unknown angles.

But not all contexts produce right-angled triangles, so we will need to develop new tools that will help us find unknown side lengths and unknown angles in these kinds of triangles. The two most important are the sine rule and the cosine rule.

In this section we begin with the sine rule, which relates the sine ratio of an angle to the opposite side in any triangle.

The sine rule

Suppose the three angles in a triangle are $A$A, $B$B and $C$C and their opposite sides have lengths $a$a, $b$b and $c$c respectively: Then the sine rule states that:

$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$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

In words, the rule states the ratio of the sine of any angle to the length of the side opposite that angle, is the same for all three angles of a triangle.

The sine rule is demonstrated below. Even though you can freely change the value of the angle $C$C, you'll notice that all three ratios stay the same. Even in the special case where $C=90^\circ$C=90° and the triangle is right-angled, each ratio remains equal to the other two.

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### Finding a side length using the sine rule

Suppose we have the angles $A$A and $B$B and the length $b$b and we want to find the length $a$a. Using the form of the sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}$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.

In this non-right-angled triangle, we have two known angles and one known side. We want to find the length of the side $PQ$PQ  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.

Using the form of the sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA=bsinB, we get:

 $\frac{PQ}{\sin48^\circ}$PQsin48°​ $=$= $\frac{18.3}{\sin27^\circ}$18.3sin27°​ Use the form of the sine rule: $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA​=bsinB​ $PQ$PQ $=$= $\frac{18.3\sin48^\circ}{\sin27^\circ}$18.3sin48°sin27°​ Multiply both sides of the equation by $\sin48^\circ$sin48°. $PQ$PQ $=$= $29.96$29.96 (to $2$2 d.p.)

Find the side length $a$a using the sine rule. ### Finding an angle using the sine rule

Suppose we know 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 $\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa=sinBb, we can solve for $A$A.

Find angle $R$R to $1$1 decimal place.  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.

 $\frac{\sin R}{28}$sinR28​ $=$= $\frac{\sin39^\circ}{41}$sin39°41​ Use the form of the sine rule $\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa​=sinBb​ $\sin R$sinR $=$= $\frac{28\times\sin39^\circ}{41}$28×sin39°41​ Multiply both sides of the equation by $28$28. $R$R $=$= $\sin^{-1}\left(\frac{28\times\sin39}{41}\right)$sin−1(28×sin3941​) Use $\sin^{-1}$sin−1 to solve for the angle. $R$R $=$= $25.5^\circ$25.5° (to $1$1 d.p.)

#### Practice question

##### question 1

Consider a triangle where two of the angles and the side included between them are known. Is there enough information to solve for the remaining sides and angle using just the sine rule? 1. Yes

A

No

B

Yes

A

No

B

##### QUESTION 2

Find the length of side $a$a using the Sine Rule.

Enter each line of working as an equation. ##### Question 3

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

Enter each line of working as an equation. ### Outcomes

#### VCMMG367 (10a)

Establish the sine, cosine and area rules for any triangle and solve related problems.