Review question

# What's this angle of elevation given two others? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R7756

## Solution

The angular elevation of the summit of a mountain is measured from three points on a straight level road. From a point due south of the summit the elevation is $\alpha$, from a point due east of it the elevation is $\beta$, and from the point of the road nearest to the summit the elevation is $\gamma$.

If the direction of the road makes angle $\theta$ east of north, prove that

1. $\tan\theta=\tan\alpha\cot\beta$

Let us choose the length $AO$ to be $1$, without loss of generality. (This says effectively that our angles will be the same whatever $AO$ is.)

Let the height of the mountain be $h$. We have three expressions for the height $h$: \begin{align*} \triangle AOM:\quad h&=\tan\alpha\\ \triangle BOM:\quad h&=\tan\theta\tan\beta\\ \triangle COM:\quad h&=\sin\theta\tan\gamma. \end{align*}

Equating $h$ from triangles $AOM$ and $BOM$, we get $h=\tan\alpha=\tan\theta\tan\beta$ so $\tan\theta=\tan\alpha\cot\beta$ as required.

1. $\tan^2\gamma=\tan^2\alpha+\tan^2\beta$.

We will use here the identity $1+\cot^2\theta=\cosec^2\theta.$

We have from the diagrams

1. $\tan^2 \gamma = \dfrac{h^2}{OC^2}$
2. $\tan^2\alpha = h^2$
3. $\tan^2\beta = \dfrac{h^2}{OB^2},$ so
\begin{align*} \tan^2\alpha + \tan^2\beta &= h^2 + \dfrac{h^2}{OB^2} \\ &=h^2\left(1 + \dfrac{1}{OB^2}\right)\\ &= h^2(1+\cot^2\theta)\\ &= h^2\cosec^2\theta\\ &= \dfrac{h^2}{\sin^2\theta}\\ &= \tan^2\gamma. \end{align*}

Find $\gamma$, if $\theta=31^{\circ}$ and $\alpha=8^{\circ}$.

We can eliminate $\beta$ from the results of (i) and (ii) to get $\tan^2\gamma=\frac{\tan^2\alpha}{\sin^2\theta}.$ Substituting in the values for $\theta$ and $\alpha$, we find $\gamma\approx15.3^\circ$.