Question

Two men on either side of a 75 m high building and in line with base of building observe the angles of elevation of the top of the building as 30° and 60°. Find the distance between the two men. (Use\[\sqrt{3} = 1 . 73\])

Answer 1

Let CD be the height of the building.
So,
CD = 75 m
It is given that two men are at points A and B and they observe angles of elevation of the top of the building as 60° and 30°, respectively.

\[In ∆ ACD, \]
\[\tan60^o = \frac{CD}{AD}\]
\[ \Rightarrow \sqrt{3} = \frac{75}{AD}\]
\[ \Rightarrow AD = 25\sqrt{3} m\]

\[In ∆ BCD, \]
\[\tan30^o = \frac{CD}{BD}\]
\[ \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{BD}\]
\[ \Rightarrow BD = 75\sqrt{3} m\]

∴ Distance between both the men = BD + AD = 

\[75\sqrt{3} + 25\sqrt{3} = 100\sqrt{3} m = 100 \times 1 . 73 = 173 m\]