UPSC NDA I and II 2020 Math Solved Paper

Concept:Use angle sum property and sine rule to verify each statement.Explanation:Statement 1: In triangle $ABC$, given $A = 2B$ and $b = c$.Since $b = c$, the angles opposite these sides are equal, so $B = C$.Also $A = 2B = 2C$.Angle sum: $A + B + C = 180^{\circ}$ ⇒ $2C + C + C = 180^{\circ}$ ⇒ $4C = 180^{\circ}$ ⇒ $C = 45^{\circ}$.Thus $B = 45^{\circ}$ and $A = 90^{\circ}$.Therefore the triangle is right-angled, not obtuse.Hence statement 1 is false.Statement 2: Given $A = 40^{\circ}$, $B = 65^{\circ}$ and $\frac{a}{c} = \sin 40^{\circ} \csc 15^{\circ}$.Using angle sum: $C = 180^{\circ} - 40^{\circ} - 65^{\circ} = 75^{\circ}$.By sine rule in $\text{triangle } ABC$, $\frac{a}{\sin A} = \frac{c}{\sin C}$ ⇒ $\frac{a}{c} = \frac{\sin A}{\sin C} = \frac{\sin 40^{\circ}}{\sin 75^{\circ}} = \sin 40^{\circ} \csc 75^{\circ}$.The given ratio is $\sin 40^{\circ} \csc 15^{\circ}$, which is different because $\csc 75^{\circ} \neq \csc 15^{\circ}$.Thus no triangle exists with the given conditions.Hence statement 2 is true.Answer:Only statement 2 is correct, so option B (2 only).