# Sine & Cosine Rules

## Sine & Cosine Rules Revision

**Sine & Cosine Rules**

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

Here’s the triangle we’ll be referencing from in this section.

**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}}

**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}

**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}

**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.

**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°

## Sine & Cosine Rules Example Questions

**Question 1:** Use the cosine rule to find the value of \alpha.

**[3 marks]**

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°

**Question 2:** For the diagram below, find an expression for the area of the triangle in terms of a.

**[2 marks]**

**Question 3:** For the diagram below, find the values of \alpha, \beta and x.

**[5 marks]**

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)}