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10 Further trigonometry
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10.05 The sine rule

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Lesson

Introduction

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

Sine rule

A triangle with angles A, B, and C with their opposite sides lengths lower case a, b, and c, respectively.

Suppose the three angles in a triangle are A, B and C and their opposite sides have lengths a, b and c respectively.

The sine rule states that \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}.

We can also take the reciprocal of each fraction to give the alternate form: \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}.

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.

Exploration

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

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As the value of C changes, the three ratios are equal. Also, when C=0 \degree and C=180 \degree then the ratios are undefined.

Examples

Example 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?

A triangle with two known angles A and B where c is a side length between angles A and B.
Worked Solution
Create a strategy

To use the sign rule we need a problem involving two side-angle pairs.

Apply the idea

Since we are given angles A and B, then we are able to find angle C through angle sum of a triangle.

So we could then find side a using the known \angle A, \angle C and side c:\dfrac{a}{\sin A}=\dfrac{c}{\sin C} Similarly for side b.

So, yes there is enough information to solve for the remaining sides and angle using the sine rule.

Idea summary

The sine rule:

\displaystyle \dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}, \quad \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}
\bm{A}
is an angle of a triangle opposite side a
\bm{B}
is an angle of a triangle opposite side b
\bm{C}
is an angle of a triangle opposite side c

Find a side length

Suppose we have the angles A and B and the length b and we want to find the length a. Using the form of the sine rule \dfrac{a}{\sin A}=\dfrac{b}{\sin B}, we can make a the subject by multiplying both sides by \sin A. This gives a=\dfrac{b\sin A}{\sin B}.

Examples

Example 2

Find the side length a using the sine rule. Round your answer to two decimal places.

A scalene triangle with side length 18 opposite an angle of 69 degrees, and side length A opposite an angle of 33 degrees.
Worked Solution
Create a strategy

Use the sine rule: \dfrac{a}{\sin A}=\dfrac{b}{\sin B}.

Apply the idea
\displaystyle \frac{a}{\sin 33\degree} \displaystyle =\displaystyle \frac{18}{\sin 69\degree }Substitute b=18, \, A=33, \, B=69
\displaystyle a\displaystyle =\displaystyle \frac{18}{\sin 69\degree} \times \sin 33\degreeMultiply both sides by \sin 33\degree
\displaystyle =\displaystyle 10.50Evaluate
Idea summary

We can find a side length using the form of the sine rule \dfrac{a}{\sin A}=\dfrac{b}{\sin B}.

Find an angle

Suppose we know the side lengths a and b and the angle B, and we want to find the angle A. Using the form of the sine rule \dfrac{\sin A}{a}=\dfrac{\sin B}{b}, we can solve for A.

Examples

Example 3

Find the value of the acute angle A using the Sine Rule. Write your answer in degrees to two decimal places.

A triangle with side length 98 opposite angle A and side length 81 opposite an angle of 50 degrees.
Worked Solution
Create a strategy

Use the sine rule: \dfrac{\sin A}{a}=\dfrac{\sin B}{b}.

Apply the idea
\displaystyle \dfrac{\sin A}{98}\displaystyle =\displaystyle \dfrac{\sin 50^\circ }{81}Substitute a=98, \, b=81, \, B=50
\displaystyle \sin A\displaystyle =\displaystyle 98 \times\frac{\sin 50\degree }{81}Multiply both sides by 98
\displaystyle \sin A\displaystyle =\displaystyle 0.926\,819Evaluate
\displaystyle A\displaystyle =\displaystyle \sin ^{-1}\left(0.926\,819\right)Take the inverse sine of both sides
\displaystyle =\displaystyle 67.94\degreeEvaluate
Idea summary

We can find an angle using the form of the sine rule \dfrac{\sin A}{a}=\dfrac{\sin B}{b}.

Outcomes

MA5.3-15MG

applies Pythagoras' theorem, trigonometric relationships, the sine rule, the cosine rule and the area rule to solve problems, including problems involving three dimensions

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