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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)

Finding a side length

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 law of sines with numerator lengths $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA=bsinB, we can solve 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 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 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.

Write your answer correct to two decimal places.

Question 3

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

Round your answer to two decimal places.

 

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

Finding a side length

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 law of sines with numerator lengths $\frac{a}{\sin A}=\frac{b}{\sin B}$asinA=bsinB, we can solve 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 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 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.

Write your answer correct to two decimal places.

Question 3

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

Round your answer to two decimal places.

 

Outcomes

G.SRT.D.10

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

G.SRT.D.11

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

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