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.
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
$\frac{\sin A}{a}$sinAa$=$=$\frac{\sin B}{b}$sinBb$=$=$\frac{\sin C}{c}$sinCc.
This can also be written as
$\frac{a}{\sin A}$asinA$=$=$\frac{b}{\sin B}$bsinB$=$=$\frac{c}{\sin C}$csinC.
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.) |
Practice question
Question 1
Find the side length $a$a using the sine rule.
Round your answer to two decimal places.
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 2
Find the value of the acute angle $x$x using the sine rule.
Round your answer to one decimal place.
A triangle is shown with vertices labeled $A$A, $B$B, and $C$C. The side opposite vertex $A$A is labeled with the length 17, and the side opposite vertex C is labeled with the length 9. The angle at vertex $A$A is given as $56^\circ$56°. The angle at vertex $C$C is labeled with the variable $x$x. The angle at vertex B is marked with an arc but is not labeled, and its opposite side, side AC, is also unlabeled.
QUESTION 3
Consider the following diagram of a quadrilateral.
Solve for $\theta$θ in degrees giving your answer correct to two decimal places.
