Question

A player sitting on the top of tower of height 20 m observes the angle of a depression of a ball lying on the ground as ${60}^{\circ }$. Find the distance between the foot of the tower and the ball. $(take\sqrt{3}=1.732)$

Answer 1

Let $AB$ be the tower and $BC$ be the distance between the foot of the tower and the ball.

As we know that $\tan \theta =\frac{\text{Perpendicular}}{\text{Base}}$

Now from fig.,

$\tan 60°=\frac{AB}{BC}$

$\Rightarrow \sqrt{3}=\frac{20}{BC}$

$\Rightarrow BC=\frac{20}{1.732}(\because \sqrt{3}=1.732)$

$\Rightarrow BC=11.547m$

Hence the distance between the foot of the tower and the ball is $11.547m$.

Hence the correct answer is $11.547m$.