Question

The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the sun is 60°. Find the angle of elevation of the sun at the time of the longer shadow ?

Answer 1

Suppose A be the position of the sun. Let the height of the tower be h m and the angle between the sun and the ground at the time of longer shadow be θ.

BC and BD are the lengths of the shadow of the tower when the angle between the sun and the ground are 60° and θ, respectively.

Given, BD = 3 BC     .....(1)

In ΔABC,

\[\tan60^o = \frac{AB}{BC}\]
\[ \Rightarrow \sqrt{3} = \frac{h}{BC}\]
\[ \Rightarrow h = \sqrt{3}BC . . . . . \left( 2 \right)\]

In ΔABD,

\[\tan\theta = \frac{AB}{BD}\]
\[ \Rightarrow \tan\theta = \frac{h}{BD}\]
\[ \Rightarrow h = \tan\theta \times BD . . . . . \left( 3 \right)\]

From (1), (2) and (3), we get

\[\sqrt{3}BC = \tan\theta \times 3BC\]

\[ \Rightarrow \tan\theta = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}\]

\[ \Rightarrow \theta = 30^o \left( \because \tan30^o= \frac{1}{\sqrt{3}} \right)\]

Thus, the angle between the sun and the ground at the time of longer shadow is 30°.