UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Use trigonometric ratios and the fact that the image of a cloud in a lake is formed at the same depth below the water surface.Explanation:Let the height of the cloud above the lake be $h$ metres.The observer is at a point $25$ m above the lake.So the vertical distance from the observer to the cloud is $(h - 25)$ m.The image of the cloud is $h$ m below the lake surface, so the vertical distance from the observer to the image is $(h + 25)$ m.Let the horizontal distance from the observer to the point directly under the cloud be $x$ m.From the angle of elevation ($15^{\circ}$) to the cloud:$\displaystyle \tan 15^{\circ} = \frac{h - 25}{x}$From the angle of depression ($45^{\circ}$) to the image:$\displaystyle \tan 45^{\circ} = \frac{h + 25}{x}$Since $\tan 45^{\circ} = 1$, we get $x = h + 25$.Substitute into the first equation:$\displaystyle \frac{h - 25}{h + 25} = \tan 15^{\circ}$Evaluate $\tan 15^{\circ}$ using $\tan(45^{\circ} - 30^{\circ})$:$\tan 15^{\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$Set up the equation:$\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{h - 25}{h + 25}$Cross‑multiply:$(\sqrt{3} - 1)(h + 25) = (\sqrt{3} + 1)(h - 25)$Expand:$\sqrt{3}h + 25\sqrt{3} - h - 25 = \sqrt{3}h - 25\sqrt{3} + h - 25$Cancel $\sqrt{3}h$ and $-25$ from both sides:$25\sqrt{3} - h = -25\sqrt{3} + h$Bring terms together:$25\sqrt{3} + 25\sqrt{3} = h + h$$50\sqrt{3} = 2h$$h = 25\sqrt{3}$ mAnswer:Option B: $25\sqrt{3}$ m