Question

The angle of elevation of the top of a hill from the foot of a tower is ${{60}^{\circ }}$ and the angle of elevation of the top of the tower from the foot of the hill is ${{30}^{\circ }}$. If the tower is 50 m high, find the height of the hill.

Answer 1

Hint: For solving this question, first let that the height of the tower be AB. And CD is the height of the hill. Now, the angle of ACB is given and angles CAD is also given. The height of AB is 50 m. Let, the height of the CD is ‘x’ m. Applying $\cot 30^{\circ}$ degrees in triangle BCA, we get the length of AC. Similarly applying $\tan 60^{\circ} $ in the triangle CAD, we get the length of the CD which is the height of the hill.

Complete step-by-step solution -
The diagrammatical representation of the situation is:

Let AB be the height of the tower and CD be the height of the hill.
Now, we are given that $\angle ACB={{30}^{\circ }},\angle CAD={{60}^{\circ }}$ and $AB=50m$.
Let the height of the CD be ‘x’ m.
In right angle $\Delta ACB,\,\text{applying cot}\theta \text{, where }\theta \text{=30}{}^\circ $:
$\begin{align}
  & \cot 30{}^\circ =\frac{AC}{AB} \\
 & \text{As, }\cot 30{}^\circ =\sqrt{3} \\
 & \therefore \sqrt{3}=\frac{AC}{50} \\
 & AC=50\sqrt{3}m \\
\end{align}$
Applying $\tan {{60}^{\circ }}$ in the right-angle $\Delta ACD$,
$\begin{align}
  & \tan 60{}^\circ =\frac{CD}{AC} \\
 & \text{As, }\tan 60{}^\circ =\sqrt{3} \\
 & \sqrt{3}=\frac{x}{50\sqrt{3}} \\
 & x=50\times 3=150m \\
\end{align}$
Hence, the height of the hill is 150m.

Note: Students must read the problem statement thoroughly to get a fair idea of diagrammatic representation of the situation. They must be familiar with the angle of elevation for a body. At last, the two triangles must be chosen wisely for applying the trigonometric ratios.