# Sine & Cosine Rules

A LevelAQAEdexcelOCRAQA 2022OCR 2022

## Sine & Cosine Rules

The two rules work for any triangle at all – not just the right angled ones we’d use usual trigonometry for.

A Level   Here’s the triangle we’ll be referencing from in this section. A Level   ## Sine Rule

Use the Sine Rule when you know the values of two angles and one side length, and want to figure out the length of another side.

The rule is

$\dfrac{\textcolor{red}{a}}{\sin \textcolor{red}{A}} = \dfrac{\textcolor{blue}{b}}{\sin \textcolor{blue}{B}} = \dfrac{\textcolor{limegreen}{c}}{\sin \textcolor{limegreen}{C}}$

A Level   ## Cosine Rule

We’ll use this rule when we know two side lengths and the angle in between. We might also use it when we know all three side lengths.

The rule is

$\textcolor{red}{a}^2 = \textcolor{blue}{b}^2 + \textcolor{limegreen}{c}^2 - 2\textcolor{blue}{b}\textcolor{limegreen}{c}\cos \textcolor{red}{A}$

A Level   ## Area of Any Triangle

This formula can be used for any pair of sides where the angle in between is also known.

The formula is

$\text{Area} = \dfrac{1}{2}\textcolor{red}{a}\textcolor{blue}{b}\sin \textcolor{limegreen}{C}$

A Level   ## Note:

If we have two sides and an angle that doesn’t lie in between, we have a bit of a problem.

We’d need to find a little more information out about the system, typically by inspecting the surrounding system.

A Level   ## Example: Application

Here’s a system of two triangles attached at the side of length $\textcolor{red}{5}\text{ cm}$.

Find values for $x$ and $\alpha$.

[4 marks] First, let’s solve for $x$.

Since we know the top triangle is isosceles, the two angles we don’t know yet are equal. Given that the total of angles in the triangle are $180°$, we have these two angles as $80°$ each.

From there

$\dfrac{\textcolor{red}{5}}{\sin 20°} = \dfrac{x}{\sin 80°}$

so

$x = \dfrac{\textcolor{red}{5}\sin 80°}{\sin 20°} = 14.40\text{ cm (to } 2 \text{ dp)}$

Now, to solve for $\alpha$.

$\textcolor{red}{5}^2 = \textcolor{blue}{6}^2 + \textcolor{limegreen}{7}^2 - (2 \times \textcolor{blue}{6} \times \textcolor{limegreen}{7} \times \cos \textcolor{red}{\alpha})$

or

$\textcolor{red}{25} = \textcolor{blue}{36} + \textcolor{limegreen}{49} - 84\cos \textcolor{red}{\alpha}$

so

$\textcolor{red}{\alpha} = \cos ^{-1}\left( \dfrac{\textcolor{red}{25} - (\textcolor{blue}{36} + \textcolor{limegreen}{49})}{-84}\right) = 44.42°$

A Level   ## Example Questions

$a^2 = b^2 + c^2 - 2bc\cos A$

gives

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

so

$A = \cos ^{-1}\left( \dfrac{b^2 + c^2 - a^2}{2bc}\right) = \cos ^{-1}\left( \dfrac{4^2 + 5^2 - 6^2}{2 \times 4 \times 5}\right) = 82.82°$

$\text{Area} = \dfrac{1}{2} \times 1.5a \times 2a \times \sin 70° = 1.41a^{2}$

First of all, we have the unmarked angle in the triangle as $180° - 120° = 60°$.

By using the sine rule, we have the equation

$\dfrac{15}{\sin 60°} = \dfrac{10}{\sin \alpha}$

giving

$\alpha = 35.26°$

Since we have angles in a triangle summing to $180°$, we have

$\beta = 180° - (60° + 35.26°) = 84.74°$

By extension,

$\dfrac{15}{\sin 60°} = \dfrac{x}{\sin \beta}$

so

$x = 17.25\text{ cm (to }2\text{ dp)}$

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