UPSC NDA II 2018 Mathematics Solved Paper

Concept:Angle of elevation is the angle measured upward from the horizontal to the line of sight.Explanation:Let the height of the hill be $H$ and the horizontal distance between the hill and the building be $x$.From the top of the building (height $h$), the angle of elevation to the hilltop is $\frac{\pi}{6}$ ($30^{\circ}$).From the bottom of the building, the angle of elevation is $\frac{\pi}{3}$ ($60^{\circ}$).In the triangle for the bottom view: $\tan 60^{\circ} = \frac{H}{x}$, so $\sqrt{3} = \frac{H}{x}$, giving $x = \frac{H}{\sqrt{3}}$.In the triangle for the top view: $\tan 30^{\circ} = \frac{H - h}{x}$, so $\frac{1}{\sqrt{3}} = \frac{H - h}{x}$.Substitute $x$: $\frac{1}{\sqrt{3}} = \frac{H - h}{\frac{H}{\sqrt{3}}}$.Simplify: $\frac{1}{\sqrt{3}} \times \frac{H}{\sqrt{3}} = H - h$, i.e., $\frac{H}{3} = H - h$.Multiply by 3: $H = 3H - 3h$, so $3h = 2H$.Thus, $H = \frac{3h}{2}$.Answer:$H = \frac{3h}{2}$, which corresponds to option B.