Question

A geologist measured a $40°$ angle of elevation to the top of a mountain. After moving 0.5 km farther away the angle of elevation was $34°$. How high is the top of the mountain?

Short answer

The top of the mountain is $1.719\text{Km}$.

Step-by-step solution

Step 1. Write down the given information.

It is given that the angle of elevation of the top of a mountain changes from $40{}^{\circ } \text{to}34°$ as the geologist moves 0.5 km farther away as shown in diagram below.

Step 2. Calculation.

Apply trigonometric ratios in triangle ABC and triangle ABD. Therefore,

$\begin{array}{c}\tan 40°=\frac{h}{x}\mathrm{....}\left(\text{From}\Delta \text{ABC}\right)\\ h=x\tan 40°\mathrm{....}\left(1\right)\\ \text{And,}\tan 34°=\frac{h}{x+0.5}\mathrm{....}\left(\text{From}\Delta \text{ABD}\right)\\ x+0.5=\frac{h}{\tan 34°}\\ x=\frac{h}{\tan 34°}-0.5\end{array}$

Plugging $x=\frac{h}{\tan 34°}-0.5$ in (1) and simplifying for $\left(h\right)$ gives,

$\begin{array}{c}h=\left(\frac{h}{\tan 34°}-0.5\right)\tan 40°\\ h=\frac{h\tan 40°}{\tan 34°}-0.5\tan 40°\end{array}$

Further simplifying,

$\begin{array}{c}\left(\frac{\tan 40°}{\tan 34°}-1\right)h=0.5\tan 40°\\ \left(\frac{\tan 40°-\tan 34°}{\tan 34°}\right)h=0.5\tan 40°\\ h=\frac{0.5\tan 40°\tan 34°}{\tan 40°-\tan 34°}\\ h=1.719\text{Km}\end{array}$

Since, the value of $h=1.719\text{Km}$. Hence, the top of the mountain is $1.719\text{Km}$.

Step 3. Conclusion.

The top of the mountain is $1.719\text{Km}$.