UPSC NDA 2 2022 Mathematics Solved Paper

Concept:Use trigonometric ratios and Pythagoras theorem to find the tower length from given distances and angles.Explanation:Let $O$ be the foot of the tower and $M$ the top.Let $A$ be a point on the ground such that $OA = a$.Points $P$ and $Q$ are at distances $x$ and $y$ from $A$, so $OP = a + x$ and $OQ = a + y$.In $\triangle MOP$, $\tan 15^\circ = \frac{OM}{OP} = \frac{m}{a+x}$.Using $\tan(45^\circ-30^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$ and rationalizing, we get $\frac{m}{a+x} = \sqrt{3} - 2$.Thus $m = (a+x)(\sqrt{3} - 2)$. …(1)In $\triangle MOQ$, $\tan 75^\circ = \frac{OM}{OQ} = \frac{m}{a+y}$.Using $\tan(45^\circ+30^\circ) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1}$ and rationalizing, we get $\frac{m}{a+y} = -(\sqrt{3}+2)$.Thus $m = -(a+y)(\sqrt{3}+2)$. …(2)Equating (1) and (2): $(a+x)(\sqrt{3} - 2) = -(a+y)(\sqrt{3}+2)$.Expand: $\sqrt{3}a - 2a + \sqrt{3}x - 2x = -\sqrt{3}a - 2a - \sqrt{3}y - 2y$.Simplify: $2\sqrt{3}a = 2(x-y) - \sqrt{3}(x+y)$.So $a = \frac{2(x-y) - \sqrt{3}(x+y)}{2\sqrt{3}}$. …(3)Substitute (3) into (1): $m = \left( \frac{2(x-y) - \sqrt{3}(x+y)}{2\sqrt{3}} + x \right)(\sqrt{3} - 2)$.Simplify inside: $\frac{2(x-y) - \sqrt{3}(x+y) + 2\sqrt{3}x}{2\sqrt{3}} = \frac{(2+\sqrt{3})(x-y)}{2\sqrt{3}}$.Then $m = \frac{(2+\sqrt{3})(x-y)}{2\sqrt{3}}(\sqrt{3} - 2)$.Since $(2+\sqrt{3})(\sqrt{3}-2) = -1$, we get $m = \frac{-(x-y)}{2\sqrt{3}}$. Taking positive length, $m = \frac{x-y}{2\sqrt{3}}$. …(4)Now in right $\triangle MOA$, $MA^2 = OM^2 + OA^2 = m^2 + a^2$.From (3) and (4): $MA^2 = \left(\frac{x-y}{2\sqrt{3}}\right)^2 + \left(\frac{2(x-y) - \sqrt{3}(x+y)}{2\sqrt{3}}\right)^2$.Factor out $\frac{(x-y)^2}{(2\sqrt{3})^2}$: $MA^2 = \frac{(x-y)^2}{(2\sqrt{3})^2} \left[ 1 + \left( \frac{2(x-y) - \sqrt{3}(x+y)}{x-y} \right)^2 \right]$.Simplify the fraction: $\frac{2(x-y) - \sqrt{3}(x+y)}{x-y} = 2 - \frac{\sqrt{3}(x+y)}{x-y}$.Thus $MA = \frac{x-y}{2\sqrt{3}} \sqrt{ 1 + \left\{ 2 - \frac{\sqrt{3}(x+y)}{x-y} \right\}^2 }$.Calculation: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: Option B: $\frac{x-y}{2 \sqrt{3}} \sqrt{1 + \left\{ 2 - \frac{\sqrt{3} (x + y)}{x - y} \right\}^2}$