Question

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° and 60° respectively. Find the height of the tower.

Answer 1

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 ∆BCD,

$\frac{BC}{CD} = tan45°$

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

$CD = 20m$

In ∆ACD,

$\frac{AC}{ CD }= tan 60°$

$\frac{AB + BC}{ CD} = \sqrt3$

$\frac{AB + 20}{ 20} = \sqrt3$

$AB = (20\sqrt3 -20)\, m$

$AB = 20(\sqrt3 -1)\,m$

Therefore, the height of the transmission tower is $20(\sqrt3 -1)\,m$.