# SOH CAH TOA

## What is SOH CAH TOA?

This phrase is your most valuable tool when solving trig problems with right-angled triangles. It summarises these three relationships:

$\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$
$\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}$
$\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$

Where "opposite", "hypotenuse" and "adjacent" refer to side lengths relative to $$\theta$$. Adjacent means "next to", and the hypotenuse is the longest side.

SOH CAH TOA comes from abbreviating Sin, Cos, Tan, Opposite, Adjacent and Hypotenuse; then arranging them like so: $\text{SOH}\Rightarrow\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}$ SOH CAH TOA helps in remembering these relations because the first letter is the trig function, equal to the second letter divided by the third. Each triplet is the same.

## When to use SOH CAH TOA

You would use these three equations in one of two situations involving right angled triangles:
- To find an angle given two side lengths.
- To find one side length given another length and one of the acute angles.

## SOH, CAH or TOA? Which to use

The trick is to look at the information you've been given or asked to find, there will always be (at least) three things. Look at where the angle is you've been given or asked to find and label the opposite and adjacent sides relative to it. Which sides do you want to use? If it's the Opposite and Hypotenuse then you want to use SOH.

## Example 1: finding $$\theta$$

\begin{align} \tan(\theta)&=\frac{\text{opposite}}{\text{adjacent}} \\[8pt] \tan(\theta)&=\frac{5}{15} \\[8pt] \tan(\theta)&=\frac{1}{3} \\[8pt] \theta&=\tan^{-1}\left(\frac{1}{3}\right) \\[8pt] \theta&=19.471^o \hspace{14pt}\text{[3 d.p.]} \end{align}

## Example 2: finding side length

\begin{align} \sin(\theta)&=\frac{\text{opposite}}{\text{hypotenuse}} \\[8pt] \sin(30)&=\frac{x}{4} \\[8pt] \frac{1}{2}&=\frac{x}{4} \\[10pt] 2&=x \end{align}