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 $20m$ high building are ${45}^{\circ }and{60}^{\circ }$ respectively. Find the height of the transmission tower.

Answer 1

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

Let, $BC$ be the building, $AB$ be the transmission tower, and $D$ be the point on the ground from where the elevation angles are to be measured.

In $\Delta BCD$,

$tan{45}^{\circ }=\frac{BC}{CD}$

$1=\frac{20}{CD}$

$\Rightarrow CD=20m$

in triangle $ACD$,

$tan{60}^{\circ }=\frac{AC}{CD}$

$\Rightarrow \sqrt{3}=\frac{AB+BC}{CD}$

$\Rightarrow \sqrt{3}=\frac{AB+20}{20}$

$\Rightarrow AB=(20\sqrt{3}-20)$

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

Hence, the height of the transmission tower is $20(\sqrt{3}-1)m$