Applications of Trigonometry

Choosing which rule or formula to use

It is important to be able to decide which Rule or Formula to use to answer a question
This table summarises the possibilities:

Sine & Cosine Rules, Area of Triangle – Harder table, IGCSE & GCSE Maths revision notes

Non-Right-Angled Triangles Diagram 2

Using the cosine rules to find angles

The Cosine Rule can be rearranged to give:

$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}$

When using the inverse cosine function (i.e. $\cos^{- 1}$ ) we can use this to find the size of angle $A$ :

$A = \cos^{- 1} (\frac{b^{2} + c^{2} - a^{2}}{2 b c})$

Make sure you can do this rearrangement, or remember this form of the formula

Using the sine rule to find angles

Sine-Rule-Ambiguous-case, IGCSE & GCSE Maths revision notes

If all we know are the lengths of $A B$ and $B C$ and the size of angle $B A C$ , there are two possible triangles that could be drawn
- one with side $B C_{1}$ (and angle $x = 102.8 °$ )
- the other with side $B C_{2}$ (and angle $y = 77.2 °$ )
- Using your calculator and the Sine Rule would only find you the possibility with angle $y$
- You may need to subtract your answer from 180° to find the other angle

Examiner Tip

In more involved exam questions, you may have to use both the Cosine Rule and the Sine Rule over several steps to find the final answer
If your calculator gives you a ‘Maths ERROR’ message when trying to find an angle using the Cosine Rule, you probably subtracted things the wrong way around when you rearranged the formula
The Sine Rule can also be written ‘flipped over’:

$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

- This is more useful when we are using the rule to find angles
- When finding angles with the Sine Rule, use the info in the question to decide whether you have the acute angle case (ie the calculator value) or the obtuse angle case (ie, minus the calculator value)
The Cosine Rule will never give you an ambiguous answer for an angle – as long as you put the right things into the calculator, the answer that comes out will be the correct angle

Worked example

In the following triangle:

General-Triangle-with-values-2, IGCSE & GCSE Maths revision notes

Find the size of angle

A B C

The three side lengths are known and we want to find an angle so use the cosine rule.

Cosine Rule: $c^{2} = a^{2} + b^{2} - 2 a b \cos C$ , where $C$ is the angle opposite side $c .$

$\begin{array}{rcl} A C^{2} & = & A B^{2} + B C^{2} - 2 (A B) (B C) \cos A B C \\ 4 . 4^{2} & = & 7 . 4^{2} + 4 . 8^{2} - 2 (7.4) (4.8) \cos A B C \end{array}$

Rearrange to make $\cos A B C$ the subject.

$\begin{array}{rcl} 4 . 4^{2} - (7 . 4^{2} + 4 . 8^{2}) & = & - 2 (7.4) (4.8) \cos A B C \\ \cos A B C & = & \frac{4 . 4^{2} - 7 . 4^{2} - 4 . 8^{2}}{- 2 (7.4) (4.8)} \end{array}$

Use your calculator to find the value of $\cos A B C$ .

$\begin{array}{rcl} \cos A B C & = & \frac{487}{592} \end{array}$

Use the cos^-1 button on your calculator to find the value of $A B C$ .

$\begin{array}{rcl} A B C & = & \cos^{- 1} (\frac{487}{592}) \\ = & 34.650542 . . . \end{array}$

$A B C = 34.7 ° (1 d . p .)$

Given that angle

A C B

is obtuse, use the Sine Rule and your answer from (a) to find the size of angle

A C B

Give your answers accurate to 1 d.p.

Substitute the values of $A B = 7.4, A C = 4.4 and A B C = 34.650542 . . . °$ into the sine rule.

Sine Rule: $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ .

$\frac{7.4}{\sin A C B} = \frac{4.4}{\sin 34.650542 . . . °}$

Rearrange the equation to find the value of $A C B .$

$\begin{array}{rcl} \sin A C B & = & \frac{7.4 \sin 34.650542 . . . °}{4.4} \\ = & 0.9562307 . . . \\ A C B & = & \sin^{- 1} (0.9562307 . . .) = 72.985502 . . . \end{array}$

Subtract from 180° to find the value of obtuse angle $A C B .$

$\begin{array}{rcl} A C B & = & 180 - 72.985502 . . . = 107.014497 . . . \end{array}$

$A C B = 107.0 ° (1 d . p .)$

Applications of Trigonometry (CIE IGCSE Maths: Extended)

Revision Note