question_answer

The angle of elevation of the top of a tower from the bottom of a building is twice that from its top. What is the height of the building, if the height of the tower is 75 m and the angle of elevation of the top of the tower from the bottom of the building is\[60{}^\circ \]?

A) 25 m                

B) 37.5 m

C) 50 m

D) 60 m

Correct Answer: C

Solution :

We have to find DC.

Given, \[2x=60{}^\circ \]

\[\therefore \]      \[x=30{}^\circ \]

In \[\Delta ABC,\]\[\tan 60{}^\circ =\frac{AB}{BC}\]

\[\Rightarrow \]   \[\frac{\sqrt{3}}{1}=\frac{75}{BC}\]

\[\therefore \]      \[BC=\frac{75}{\sqrt{3}}=\frac{75\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\]

\[=25\sqrt{3}\,\,cm\]

In \[\Delta AED,\]\[\tan 30{}^\circ =\frac{AE}{ED}=\frac{AE}{25\sqrt{3}}\]         \[[\because BC=ED]\]

\[\Rightarrow \]   \[\frac{1}{\sqrt{3}}=\frac{AE}{25\sqrt{3}}\]

\[\therefore \]      \[AE=25\,\,m\]

\[\therefore \]\[DC=EB=AB-AE=75-25=50\,\,cm\]