Two persons are on either sides of a tower of height 50 m. The persons observers the top of the tower at an angle of elevation of 30° and 60°. If a car crosses these two persons in 10 seconds, what is the speed of the car?

Answer: Option A

Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, ∠ BAD = 30°, ∠ BCD = 60°
$$\begin{array}{rl}& \text{From}\text{the}\text{right}\Delta ABD,\\ & \tan 30^{\circ }=\frac{BD}{BA}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{50}{BA}\\ & \Rightarrow BA=50\sqrt{3}\\ & \text{From}\text{the}\text{right}\Delta CBD,\\ & \tan 60^{\circ }=\frac{BD}{BC}\\ & \Rightarrow \sqrt{3}=\frac{50}{BC}\\ & \Rightarrow BC=\frac{50}{\sqrt{3}}\\ & =\frac{50\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\ & =\frac{50\sqrt{3}}{3}\end{array}$$
Distance between the two persons
$$\begin{array}{rl}& =AC=BA+BC\\ & =50\sqrt{3}+\frac{50\sqrt{3}}{3}\\ & =\sqrt{3}\left(50+\frac{50}{3}\right)\\ & =\frac{200\sqrt{3}}{3}\text{m}\end{array}$$
i.e., the distance travelled by the car in 10 seconds = $$\frac{200\sqrt{3}}{3}\text{m}$$
$$\begin{array}{rl}& \text{Speed}\text{of the car}=\frac{\text{Distance}}{\text{Time}}\\ & =\frac{\left(\frac{200\sqrt{3}}{3}\right)}{10}=\frac{200\sqrt{3}}{3}\text{m/s}\\ & =\frac{200\sqrt{3}}{3}\times \frac{18}{5}\text{km/hr}\\ & =\text{24}\sqrt{3}\text{km/hr}\end{array}$$