UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use the formula for $\tan(\alpha + \theta)$ to relate given distances and solve for $\tan\theta$.Explanation:Let $\alpha = \tan^{-1}(0.5)$, so $\tan\alpha = \frac{1}{2}$.From triangle $\text{triangle PAB}$, let base $x$ and perpendicular $10$, so $\tan\theta = \frac{10}{x}$. ----(1)From triangle $\text{triangle PAC}$, $\tan(\alpha+\theta) = \frac{20+10}{x} = \frac{30}{x}$.Apply the identity: $\tan(\alpha+\theta) = \frac{\tan\alpha + \tan\theta}{1 - \tan\alpha \tan\theta}$.Thus $\frac{ \frac{1}{2} + \frac{10}{x} }{ 1 - \frac{1}{2} \cdot \frac{10}{x} } = \frac{30}{x}$.Simplify numerator: $\frac{x+20}{2x}$; denominator: $\frac{2x-10}{2x}$.So $\frac{x+20}{2x-10} = \frac{30}{x}$.Cross-multiply: $x(x+20) = 30(2x-10)$ => $x^2 + 20x = 60x - 300$.Rearrange: $x^2 - 40x + 300 = 0$.Factorise: $(x-10)(x-30) = 0$, so $x = 10$ or $x = 30$.From (1), $\tan\theta = \frac{10}{10}$ or $\frac{10}{30}$, i.e., $1$ or $\frac{1}{3}$.Among the given options, $\frac{1}{3}$ is possible.Answer:$\frac{1}{3}$ (Option C).