# 7.02 The 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?

Once we move away from right-angled triangles, the trigonometric ratios do not hold. However, by forming right-angled triangles by dropping a perpendicular line from a vertex to the opposite side, we can find new ways 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 trigonometric ratios 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.

Rearranging both of these equations to make $x$x the subject will allow us 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

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.

Try proving the last part of this formula. Does this still hold for obtuse angled and right-angled triangles?

## 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 the lengths in the numerator $\frac{a}{\sin A}=\frac{b}{\sin B}$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

##### Example 1

Find the length of $PQ$PQ to $2$2 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°​ 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°. $=$= $29.96$29.96 (to $2$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 the sines in the numerator $\frac{\sin A}{a}=\frac{\sin B}{b}$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).

Caution

When finding an angle using the sine rule there may be two possibilities:

If $A$A is acute: $A=\sin^{-1}\left(\frac{a\sin B}{b}\right)$A=sin1(asinBb)

or

If $A$A is obtuse: $A=180^\circ-\sin^{-1}\left(\frac{a\sin B}{b}\right)$A=180°sin1(asinBb)

Using inverse sine on our calculators we will find the acute possibility and the obtuse possibility can be found by subtracting the acute angle from $180^\circ$180°. For more detail see 'the ambiguous case' below.

#### Worked example

##### Example 2

Find angle $R$R to $1$1 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. We can also see in the triangle that we are finding an acute angle.

Do:

 $\frac{\sin R}{28}$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. $=$= $25.5^\circ$25.5° (to $1$1 d.p.)

#### Practice questions

##### Question 1

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

##### Question 2

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

## The ambiguous case

Sometimes we are not given enough information to uniquely define a triangle. For the sine rule when finding an unknown angle it may be the case that two different triangles can be formed using the same information. If both are possible under the given conditions then we call this situation the ambiguous case

Let's start by investigating this applet and try to observe what conditions lead to there being two possible triangles.

Video instructions can be found here:

 Created with Geogebra
 Set the value of the length of $a$a (blue side), and the length of $b$b (red side). Click "Show triangle" and move the point $A$A to change the value of $\angle CAB$∠CAB (blue angle). These represent the three known quantities - two lengths and an angle. The angle $\angle ABC$∠ABC (red angle) is the value we find with the sine rule. You will sometimes be able to see a second triangle, with an additional point $B'$B′ and another possible solution $\angle AB'C$∠AB′C (green angle). What do you notice about the relationship between these two solutions?

There are two cases, and what separates one from the other is summarised in this table.

 When the length of the (blue) side opposite the known angle is equal to or greater than the length of the other known (red) side, there is only one possible triangle, and only one possible value for the other angle. When the length of the (blue) side opposite the known angle is less than the length of the other known (red) side, there are two possible triangles, and two possible values for the other angle. This "second triangle" is shown here on its own.

So when finding an angle using the sine rule, look for additional information in the question that may indicate whether an acute or obtuse angle is required, this information may be implied by use of a diagram. If no additional information is given and the length of the side opposite the known angle (matching letter) is smaller than the other known side, state both angles as possible solutions.

#### Worked example

##### Example 3

Suppose we are given $A=43^\circ$A=43°, $a=11$a=11 cm and $b=15.5$b=15.5 cm. We wish to determine the angle $B=\angle ABC$B=ABC opposite the side of length $b$b, to $1$1 decimal place.

Think: We do not know whether the triangle has an obtuse angle or not - the question has not given us additional information. And we need to be mindful that we are in the ambiguous case, as the value of $a$a is smaller than the value of $b$b. According to the sine rule,

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

We can use this rule to find $B$B since we know the values of $a$a, $b$b, and $A$A

Do: Substituting in the values provided into the version of the sine rule above gives us this equation:

$\frac{\sin B}{15.5}=\frac{\sin43^\circ}{11}$sinB15.5=sin43°11.

Rearranging,
 $\sin B$sinB $=$= $15.5\times\frac{\sin43^\circ}{11}$15.5×sin43°11​ $=$= $0.961$0.961

Using the inverse sine function, we calculate:

$B=\sin^{-1}(0.961)=73.9^\circ$B=sin1(0.961)=73.9° or $B=180^\circ-\sin^{-1}(0.961)=106.1^\circ$B=180°sin1(0.961)=106.1°

This means that with the information we were given, there are two possible triangles that can be formed, one with $B=73.9^\circ$B=73.9° and one with $B=106.1^\circ$B=106.1°.

#### Practice questions

##### Question 3

Consider $\triangle ABC$ABC below:

1. Find $x$x, noting that $x$x is acute.

2. Now find $\angle ADB$ADB to the nearest whole degree, given that$\angle ADB>\angle ACB$ADB>ACB.

##### Question 4

$\triangle ABC$ABC consists of angles $A$A, $B$B and $C$C which appear opposite sides $a$a, $b$b and $c$c respectively where $\angle CAB=36^\circ$CAB=36°, $a=7$a=7 and $b=10$b=10.

1. Select the most appropriate option to complete the sentence below:

The triangle:

Can be either acute or obtuse.

A

Must be an obtuse triangle.

B

Must be an acute triangle.

C

Does not exist.

D

Must be a right-angled triangle.

E

Can be either acute or obtuse.

A

Must be an obtuse triangle.

B

Must be an acute triangle.

C

Does not exist.

D

Must be a right-angled triangle.

E

### Outcomes

#### 4.2.1.3

establish and use the sine (ambiguous case is required) and cosine rules and the formula 𝐴rea=1/2 bc sin𝐴 for the area of a triangle

#### 4.2.1.4

construct mathematical models using the sine and cosine rules in two- and three-dimensional contexts (including bearings in two-dimensional context) and use the model to solve problems; verify and evaluate the usefulness of the model using qualitative statements and quantitative analysis