What's this angle of elevation given two others?

Question

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

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

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

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

Suggestion

Before starting, it’s always a good idea to sketch out the problem. Here, we might do that as if we’re looking down on it. We could define $h$ as the height of the mountain, and then draw three more triangles with the elevation labelled.

Can you use the three different elevation triangles to get three different expressions for the mountain’s height $h$ ?

Are there any useful trigonometrical identities that we can use?

Solution

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.

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

We have from the diagrams

$\tan^2 \gamma = \dfrac{h^2}{OC^2}$

$\tan^2\alpha = h^2$

$\tan^2\beta = \dfrac{h^2}{OB^2},$ 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\cos ec^2\theta\\ &= \dfrac{h^2}{\sin^2\theta}\\ &= \tan^2\gamma. \end{align*}

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 $\gamma\approx15.3^\circ$ .