Question

A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point $A$ on the path, he observes that the angle of elevation of the top of the pillar is ${30}^{\circ }$. After walking for $10$ minutes from $A$ in the same direction, at a point $B,$ he observes that the angle of elevation of the top of the pillar is ${60}^{\circ }$. Then the time taken (in minutes) by him, from $B$ to reach the pillar is:

• A. $6$
• B. $10$
• C. $20$
• D. $5$

Answer 1

The correct option is D $5$

In $△BCD,$

$\tan {60}^{\circ }=\frac{h}{y}$

$\Rightarrow \sqrt{3}=\frac{h}{y}$

$\Rightarrow h=\sqrt{3}y\dots \left(i\right)$

In $△ACD,$

$\begin{array}{cc}\tan {30}^{\circ }& =\frac{h}{x+y}\\ \frac{1}{\sqrt{3}}& =\frac{\sqrt{3}y}{x+y}\dots \left[\text{By using}\right(i\left)\right]\\ x+y& =3y\\ x& =2y\\ y& =\frac{x}{2}\end{array}$

A man walks $x$ distance in $10$ minutes.

So, the time taken by man to walk $\frac{x}{2}$ distance will be $5$ minutes.