question_answer

From the top of a building 60 m high the angles of depression of the top and bottom of a tower are observed to be \[30{}^\circ \]and \[60{}^\circ .\]Then, the height of the tower is

A)  10 m               

B)  20 m   

C)  30 m                           

D)  40 m

Correct Answer: D

Solution :

Let x be the height of the tower.

AB = 60m, BD = y m

In \[\Delta ABD,\]

\[\frac{AB}{BD}=\tan 60{}^\circ \]or\[\frac{60}{y}=\sqrt{3}\]\[\Rightarrow \]\[y=\frac{60}{\sqrt{3}}\]       ?(i)

In\[\Delta AEC,\]\[\frac{AE}{EC}=\tan 30{}^\circ \]

\[\Rightarrow \]   \[\frac{AE}{BD}=\frac{1}{\sqrt{3}}\]\[\Rightarrow \]\[\frac{AE}{Y}=\frac{1}{\sqrt{3}}\]\[\Rightarrow \]\[\frac{AE}{\frac{60}{\sqrt{3}}}=\frac{1}{\sqrt{3}}\]

\[AE=\frac{1}{\sqrt{3}}\times \frac{60}{\sqrt{3}}=20\,\,m\]

\[\therefore \]      \[x=AB-AE=60-20=40\,\,m\]

So, height of tower = 40 m