UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Using trigonometric cotangent in right triangles formed by the tower and two observation points, and subtracting equations to eliminate the common distance from the foot.Explanation:Let the tower height be $h$ and the common distance from the foot of the tower to the point directly below the top be $a$.For the first station (distance $x + a$ from foot), angle of elevation $15^{\circ}$ gives $\cot 15^{\circ} = \frac{x + a}{h}$. Thus $x + a = h\cot 15^{\circ}$. Multiply by $y$: $xy + ay = hy\cot 15^{\circ}$ ... (1).For the second station (distance $y + a$ from foot), angle of elevation $75^{\circ}$ gives $\cot 75^{\circ} = \frac{y + a}{h}$. Thus $y + a = h\cot 75^{\circ}$. Multiply by $x$: $xy + ax = hx\cot 75^{\circ}$ ... (2).For the third point at distance $a$ from foot, $\cot \theta = \frac{a}{h}$ ... (3).Subtract (1) from (2): $a(x - y) = h(x\cot 75^{\circ} - y\cot 15^{\circ})$. Then $\frac{a}{h} = \frac{x\cot 75^{\circ} - y\cot 15^{\circ}}{x - y}$, i.e., $\cot \theta = \frac{x\cot 75^{\circ} - y\cot 15^{\circ}}{x - y}$.Substitute $\cot 75^{\circ} = 2 - \sqrt{3}$ and $\cot 15^{\circ} = 2 + \sqrt{3}$: $\cot \theta = \frac{x(2 - \sqrt{3}) - y(2 + \sqrt{3})}{x - y} = \frac{2x - \sqrt{3}x - 2y - \sqrt{3}y}{x - y} = \frac{2(x - y) - \sqrt{3}(x + y)}{x - y} = 2 - \frac{\sqrt{3}(x + y)}{x - y}$.Answer:$2 - \frac{\sqrt{3}(x + y)}{x - y}$ (Option D)