UPSC NDA 2 2017 Mathematics Solved Paper

Concept:Use $\tan \theta$ to relate heights and horizontal distance from angles of elevation.Explanation:Let the horizontal distance between the pole and the tower be $x$.From the top of the pole, angle of elevation to tower top is $30^{\circ}$: $\tan 30^{\circ} = \frac{h_T - h_P}{x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h_T - h_P}{x} \Rightarrow x = \sqrt{3}(h_T - h_P)$. (1)From the foot of the pole, angle of elevation is $45^{\circ}$: $\tan 45^{\circ} = \frac{h_T}{x} \Rightarrow 1 = \frac{h_T}{x} \Rightarrow x = h_T$. (2)Equate (1) and (2): $\sqrt{3}(h_T - h_P) = h_T$. Simplify to $\sqrt{3}h_T - \sqrt{3}h_P = h_T \Rightarrow \sqrt{3}h_P = h_T(\sqrt{3} - 1)$. Thus $\frac{h_P}{h_T} = \frac{\sqrt{3} - 1}{\sqrt{3}} = \frac{3 - \sqrt{3}}{3}$. Hence $h_T = \frac{3h_P}{3 - \sqrt{3}}$.Now check each statement using this relation.Statement 1: $\frac{2 h_P h_T}{3 + \sqrt{3}} = h_P^2$? Substitute $h_T$: left side becomes $h_P^2$. So statement 1 is correct.Statement 2: $\frac{h_T - h_P}{\sqrt{3} + 1} = \frac{h_P}{2}$? Substitute $h_T$ and simplify to get equality. So statement 2 is correct.Statement 3: $\frac{2(h_P + h_T)}{h_P} = 4 + \sqrt{3}$? Simplify left side: $2 + 2\frac{h_T}{h_P} = 2 + 2\cdot\frac{3}{3 - \sqrt{3}} = \frac{6}{3 - \sqrt{3}}$, which is not equal to $4 + \sqrt{3}$. So statement 3 is incorrect.Answer:C. 1 and 2 only