UPSC NDA II 2024 Mathematics Paper

Concept:Using sum and product of roots to find $\tan(\alpha+\beta)$ and then the third angle of the triangle.Explanation:Given that $\tan\alpha$ and $\tan\beta$ are the roots of $7x^{2} - 6x + 1 = 0$.Sum of roots: $\tan\alpha + \tan\beta = \frac{6}{7}$.Product of roots: $\tan\alpha \cdot \tan\beta = \frac{1}{7}$.Now, $\tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} = \frac{6/7}{1 - 1/7} = 1$.Thus $\alpha+\beta = 45^{\circ}$.Since the angles of the triangle are $2\alpha$, $2\beta$, and the third angle, we have $2\alpha + 2\beta = 90^{\circ}$.Therefore, the third angle is $180^{\circ} - 90^{\circ} = 90^{\circ}$, so the triangle is right‑angled.Now check if it is isosceles: compute $\tan\alpha - \tan\beta = \sqrt{(\tan\alpha+\tan\beta)^2 - 4\tan\alpha\tan\beta} = \frac{2\sqrt{2}}{7}$.Then $\tan(\alpha-\beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta} = \frac{1}{2\sqrt{2}}$.Further, $\tan 2(\alpha-\beta) = \frac{2\tan(\alpha-\beta)}{1 - \tan^{2}(\alpha-\beta)} = \frac{4\sqrt{2}}{7}$ (non‑zero).This implies $2\alpha \neq 2\beta$, so the triangle is not isosceles.Answer:The triangle is right‑angled but not isosceles, hence option (C) is correct.