Question

A tower vertically on the ground. At a point on the ground, 15m away from the foot of the tower, the angle of elevation of the top of the tower is ${{60}^{\circ }}$. What is the height of the tower?

Answer 1

Hint: Now to solve the given figure we will first draw the right angle triangle with the tower and the given point. Now we are given that the angle of elevation is ${{60}^{\circ }}$. Hence we will take tan ratio of this angle and find the value of height of the tower.

Complete step by step solution:
Now let us say the height of the tower is h. Let OL be the tower where O is the foot of the tower. Now we know that the tower will stand at right angles to the ground. Hence we get the $\angle LOA={{90}^{\circ }}$ . Now let point A be 15m away from the foot of the tower. Hence we have OA = 15m. Now the angle of elevation of tower from A is ${{60}^{\circ }}$ . Now let us first draw the figure of the given conditions.

Now consider the triangle $\Delta OAL$.
Now we know that in a right angle triangle $\tan \theta =\dfrac{\text{opposite side}}{\text{adjacent side}}$.
Now consider the ratio tan for the angle $\angle OAL$.
Hence we get $\tan \angle OAL=\dfrac{OL}{OA}$
$\begin{align}
  & \Rightarrow \tan {{60}^{\circ }}=\dfrac{OL}{OA} \\
 & \Rightarrow \sqrt{3}=\dfrac{h}{15} \\
 & \Rightarrow h=15\sqrt{3} \\
\end{align}$
Hence the height of the tower is $15\sqrt{3}m$.

Note: Now note that here we wanted to find the height of the triangle which was opposite to the given angle and we were given an adjacent side of the angle known. Now there are only two trigonometric ratios which relate opposite sides and adjacent sides which are tan and cot. Hence we can use any of the ratios to solve the problem.