Question

From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of 20 m high building are ${45}^{\circ }$ and ${60}^{\circ }$ respectively. Find the height of the tower ?

• A. $20\sqrt{3}m$
• B. 60 m
• C. $20(\sqrt{3}-1)m$
• D. $40\sqrt{3}m$

Answer 1

Answer: (C)

$20(\sqrt{3}-1)m$

Let PQ be the building of height $20$ metres,
AP be the transmission tower of height $x$ metres.
Let B be the point from where angles of elevation are drawn
$\angle PBQ={45}^{\circ },\angle ABQ={60}^{\circ }$
$In\Delta PBQ,\frac{PQ}{BQ}=\tan {45}^{\circ }\Rightarrow BQ=PQ=20m$
$In\Delta ABQ,\frac{AQ}{BQ}=\tan {60}^{\circ }\Rightarrow \frac{x+20}{20}=\sqrt{3}\Rightarrow x+20=20\sqrt{3}\Rightarrow x=20\sqrt{3}-20=20(\sqrt{3}-1)$ metres