Question

Question 7
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are ${45}^{\circ }$ and ${60}^{\circ }$ respectively. Find the height of the tower.

Answer 1

Let BC be the 20 m high building.
D is the point on the ground from where the elevation is taken.
Height of transmission tower $=AB=AC-BC$
As per question,
In right $\Delta BCD$,
tan ${45}^{\circ }=\frac{BC}{CD}$

$\Rightarrow 1=\frac{20}{CD}$
$\Rightarrow CD=20m$
Also,
In right $\Delta ACD$,
tan ${60}^{\circ }=\frac{AC}{CD}$
$\Rightarrow \sqrt{3}=\frac{AC}{20}$
$\Rightarrow AC=20\sqrt{3}m$
Height of transmission tower $=AB=AC-BC=(20\sqrt{3}-20)m=20(\sqrt{3}-1)m.$