# 8.08 Law of sines

Lesson

If we have a right 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 law of sines

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 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 law of sines.

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 law of sines shows that the lengths of the sides in a triangle are proportional to the sines of the angles opposite them.

More generally, we can use the law of sines to solve a triangle if we know either of the following:

• The measure of two angles and a side opposite one of the angles (AAS)
• The measure of two angles and a side between them (ASA)
• The measures of two sides and the angle opposite one of the sides (SSA)

## Solving a triangle given two angles and a side (AAS and ASA)

When we are given two of the three angles in a triangle, we can use the Triangle Sum Theorem to find the size of the remaining angle. To find an unknown side length, we can make use of the law of sines as follows:

### Finding a side length

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 law of sines with lengths in the numerator, $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA=bsinB, we can rearrange this for $a$a by multiplying both sides by $\sin A$sinA. This gives

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

If we are given the measure of two angles and a side between them, suppose angles $A$A and $B$B and the length $c$c, we have one extra step to do before we can proceed as above. Since the law of sines relates to opposite side-angle pairs, we need the measure of angle $C$C. We can use the Triangle Sum Theorem to calculate $m\angle C$mC and then we are set to find the lengths $a$a or $b$b.

#### Worked example

##### Question 1

Solve: Find the length $PQ$PQ correct 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 pair. This means we can use the law of sines.

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$2 d.p.)

#### Practice questions

##### Question 2

Find the length of side $a$a using the Law Of Sines.

##### Question 3

Find the side length $a$a using the law of sines.

## Solving a triangle given two sides and a non-included angle (SSA)

### Finding an angle

Suppose we have the side lengths $a$a and $b$b and angle $B$B which is opposite one of the sides. We can use this to find the angle $A$A opposite the other known side length, by using the form of the law of sines with angles in the numerator, $\frac{\sin A}{a}=\frac{\sin B}{b}$sinAa=sinBb. We can rearrange this for $\sin A$sinA by multiplying both sides by $a$a, which gives

$\sin(A)=\frac{a\sin B}{b}$sin(A)=asinBb.

From here, we can use the inverse sin function on our calculator to find the measure of $A$A, but there are two cases depending on whether $A$A is acute or obtuse in the triangle.

• If angle $A$A is acute, the inverse sin function will give us the correct value: $A=\sin^{-1}\left(\frac{a\sin B}{b}\right)$A=sin1(asinBb).
• If instead angle $A$A is obtuse, we subtract the inverse sin value from $180^\circ$180° to find the correct value: $A=180^\circ-\sin^{-1}\left(\frac{a\sin B}{b}\right)$A=180°sin1(asinBb)

Once we know the values of two sides and two angles, we can proceed to find the remaining angle by using the Triangle Sum Theorem, and the remaining side by the method given above (in the AAS and ASA section).

#### Worked example

##### Question 4

Solve: Find the value of $\angle R$R, rounding to the nearest degree.

Think: The angle we want to find is opposite a known side, and we also know a matching side and angle pair. This means we can use the law of sines. We can also see that the angle is acute, based on the other given information.

Do:

 $\frac{\sin R}{28}$sinR28​ $=$= $\frac{\sin39^\circ}{41}$sin39°41​ $R$R $=$= $\sin^{-1}\left(\frac{28\sin39^\circ}{41}\right)$sin−1(28sin39°41​), since the angle is acute $R$R $=$= $25^\circ$25°, to the nearest degree

#### Practice questions

##### Question 5

Find the value of the acute angle $x$x using the law of sines.

##### Question 6

Find the value of $x$x in degrees, given that $x$x is obtuse.

### Outcomes

#### G-SRT.10

(+) Prove the laws of sines and cosines and use them to solve problems.

#### G-SRT.11

(+) Understand and apply the law of sines and the law of cosines to find unknown measurements in right and non-right triangles.