UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Use trigonometric ratios and the tangent sum/difference formulas to relate the height of the tower to given distances and angles of elevation.Explanation:Let the tower's top be M and its foot be O, so OM = $h$.Let A be a reference point on the ground at a distance $a$ from O.Points P and Q are such that $AP = x$ and $AQ = y$, and the angles of elevation from P and Q to M are $15^{\circ}$ and $75^{\circ}$ respectively.Then $OP = a + x$, $OQ = a + y$.In triangle MOP: $\tan 15^{\circ} = \frac{h}{a+x}$.Using $\tan 15^{\circ} = \tan(45^{\circ} - 30^{\circ}) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$. Rationalising gives $\tan 15^{\circ} = \sqrt{3} - 2$.Thus $h = (a+x)(\sqrt{3} - 2)$. …(1)In triangle MOQ: $\tan 75^{\circ} = \frac{h}{a+y}$.Using $\tan 75^{\circ} = \tan(45^{\circ} + 30^{\circ}) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$. Rationalising gives $\tan 75^{\circ} = -(\sqrt{3}+2)$.Thus $h = -(a+y)(\sqrt{3}+2)$. …(2)Equating (1) and (2): $(a+x)(\sqrt{3}-2) = -(a+y)(\sqrt{3}+2)$.Expanding and solving for $a$: $2\sqrt{3}a = 2(x-y) - \sqrt{3}(x+y)$ ⇒ $a = \frac{2(x-y) - \sqrt{3}(x+y)}{2\sqrt{3}}$.Substitute $a$ into (1) and simplify:$h = \left(\frac{2(x-y) - \sqrt{3}(x+y)}{2\sqrt{3}} + x\right)(\sqrt{3}-2) = \frac{(x-y)(2+\sqrt{3})}{2\sqrt{3}}(\sqrt{3}-2)$.Since $(2+\sqrt{3})(\sqrt{3}-2) = -1$, we get $h = \frac{x-y}{2\sqrt{3}}$.Application:According to the question, the tower leaning towards north and has two points P and Q subtending an angle of $15^{\circ}$ and $45^{\circ}$ respectively(as can be seen in the figure)Answer:$h = \frac{x-y}{2\sqrt{3}}$, which corresponds to option A.