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 \(20(\sqrt3 -1)\,m\).