Question

From a point on the ground, the angle of elevation of the top of a tower is observed to be $60$ degrees. From a point, $40m$vertically above the first point of observation, the angle of elevation of the top of the tower is $30$ degrees. The horizontal distance from the point of observation is

• A. $–20\surd 3m$
• B. $\frac{\surd 3}{20}m$
• C. $–\frac{\surd 3}{20}m$
• D. $20\surd 3m$

Answer 1

The correct option is D

$20\surd 3m$

Explanation for the correct option:

Step 1. Draw the figure according to the question:

Let Point B is $40$ meter above point A

From the figure, we have

$AB=40m$

$\angle QP=60°$

$\angle QBR=30°$

$AB=PR$

$=40m$

$AP=BR$

Step 2. In $△QAP$, we have

$\tan 60°=\frac{PQ}{QP}$

$\sqrt{3}=\frac{PR+QR}{AP}$

$\sqrt{3}=\frac{40+QR}{AP}$

$\sqrt{3}AP=40+QR$ ……(1)

Step 3. In $△QBR$, we have

$\tan 30°=\frac{QR}{BR}$

$\frac{1}{\sqrt{3}}=\frac{QR}{AP}$

$AP=\sqrt{3}QR$ ……(2)

Step 4. Put the value of $AP$ in equation (1)

$\sqrt{3}\left(\sqrt{3}QR\right)=40+QR$

$3QR-QR=40$

$2QR=40$

$QR=20$

$\because$Length of tower $\left(PQ\right)=PR+QR$

$=40+20$

$=60m$

$\therefore$Distance of tower from the feet of point $A=AP$

$=\sqrt{3}QR$

$=\sqrt{3}\times 20$

$=\mathbf{20}\sqrt{\mathbf{3}}\mathbf{}\mathit{m}$

Hence, Option ‘D’ is Correct.