question_answer

From the top of a 60 m high tower, the angle of depression of the top and the bottom of a building are observed to be \[30{}^\circ \] and \[60{}^\circ \] respectively. The height of the building is                                                [WBSSC (CGL) Pre 2014]

A) \[60\sqrt{3}\]                

B) \[40\sqrt{3}\]

C) 40 m    

D) 20 m

Correct Answer: C

Solution :

Let DC is the tower and AB is the building.

In \[\Delta ADE,\]\[\tan 30{}^\circ =\frac{60-h}{x}\]

\[\Rightarrow \]   \[\frac{1}{\sqrt{3}}=\frac{(60-h)}{x}\]

\[\Rightarrow \]   \[x=(60-h)\sqrt{3}\]                    ... (i)

Now, in \[\Delta BDC,\]

\[\tan 60{}^\circ =\frac{60}{x}\] \[\Rightarrow \]   \[\sqrt{3}=\frac{60}{x}\]

\[\Rightarrow \]   \[x\sqrt{3}=60\]

\[\Rightarrow \]\[[(60-h)\sqrt{3}]\,\sqrt{3}=60\]                  [from Eq.(i)]

\[\Rightarrow \]   \[(60-h)\times 3=60\]

\[\Rightarrow \]   \[60-h=20\]

\[\Rightarrow \]   \[h=60-20=40\]

\[\therefore \]      \[h=40\,m\]

So, height of the building is 40 m.