SOHCAHTOA
Definition
The SOHCAHTOA method is used to find a side or angle in a right-angled triangle.
The longest side of the right-angled triangle is called the hypotenuse. The side opposite the angle we are using is labelled opposite, and the remaining side next to the angle is labelled adjacent.
The mnemonic SOHCAHTOA refers to which trig function should be used depending on the information given:
\begin{align} \text{SOH } &\longrightarrow & \sin\theta &= \dfrac{\text{opposite} }{\text{hypotenuse} } = \frac{\mathrm{O} }{\mathrm{H} } \\ \text{CAH } &\longrightarrow & \cos\theta &= \dfrac{\text{adjacent} }{\text{hypotenuse} } = \frac{\mathrm{A} }{\mathrm{H} } \\ \text{TOA } &\longrightarrow & \tan\theta &= \dfrac{\text{opposite} }{\text{adjacent} } = \frac{\mathrm{O} }{\mathrm{A} } \end{align}
Worked Examples
Example 1
In a right-angled triangle, find the angle $\theta$ given that the side opposite is $4$ and the hypotenuse is $6$.
Solution
The given sides are the Opposite and Hypotenuse, so use $\sin \theta = \dfrac{O}{H}$.
\begin{align} \sin \theta &= \frac{4}{6}\\ \theta &= \sin^{-1} \left(\frac{2}{3}\right)\\ \theta &= 41.8^{\circ} \text{ (to 3 sig.fig.)} \end{align}
Example 2
In a right-angled triangle, find the angle $\theta$ given the side opposite is $7$ and the adjacent is $3$.
Solution
The given lengths are the Opposite and Adjacent sides, so use $\tan \theta = \dfrac{O}{A}$.
\begin{align} \tan \theta &= \frac{7}{3}\\ \theta &= \tan^{-1} \left(\frac{7}{3}\right)\\ \theta &= 66.8^{\circ} \text{ (to 3 sig.fig.)} \end{align}
Example 3
In a right angled triangle, find the length of the hypotenuse, given $\theta = 55^{\circ}$ and the adjacent is $3$.
Solution
The given sides are the Hypotenuse and Adjacent, so use $\cos \theta = \dfrac{A}{H}$.
\begin{align} \cos 55 &= \frac{3}{H}\\\\ H \cos 55 &= 3\\ H &= \frac{3}{\cos 55}\\\\ H &= 5.23 \text{ (to 3 sig.fig.)} \end{align}
Example 4
In a right angled triangle of hypotenuse $5$, find the length of the side opposite the angle $\theta = 23^{\circ}$.
Solution
The sides given are the Opposite and the Hypotensue, so use $\sin \theta = \dfrac{O}{H}$.
\begin{align} \sin 23 &= \frac{O}{5}\\\\ 5 \sin 23 &= O\\\\ O &= 1.95 \text{ (to 3 sig.fig.)} \end{align}
External Resources
- Trigonometrical ratios in a right-angled triangle workbook at mathcentre.
