UPSC CDS 1 2021 Math Paper

Concept:Use tangent ratios with the tower height and side lengths, then apply Pythagoras to find the diagonal distance.Explanation:Let the tower be at one corner of the rectangle and its height be $h$.Let the distances along the sides to the two nearer corners be $a$ and $b$.At the first nearer corner, the angle is $60^{\circ}$.So $\tan 60^{\circ} = \frac{h}{a} \Rightarrow \sqrt{3} = \frac{h}{a} \Rightarrow a = \frac{h}{\sqrt{3}}$.At the second nearer corner, the angle is $45^{\circ}$.So $\tan 45^{\circ} = \frac{h}{b} \Rightarrow 1 = \frac{h}{b} \Rightarrow b = h$.The farthest corner is diagonally opposite the tower.Its distance from the tower is $\sqrt{a^{2} + b^{2}} = \sqrt{\frac{h^{2}}{3} + h^{2}} = h\sqrt{\frac{4}{3}} = \frac{2h}{\sqrt{3}}$.At this farthest corner, the angle subtended by the tower is $\theta$.Thus $\tan\theta = \frac{h}{\frac{2h}{\sqrt{3}}} = \frac{\sqrt{3}}{2}$.Therefore $\cot\theta = \frac{1}{\tan\theta} = \frac{2}{\sqrt{3}}$.Answer:$\cot\theta = \frac{2}{\sqrt{3}}$ (Option C).