UPSC NDA I 2023 Mathematics Solved Paper

Concept:Use the given relation $a^2+b^2+c^2 = ac + \sqrt{3}\,bc$ along with the Cosine Rule to determine the angles, then apply trigonometric ratios to find the sides and compute the area.Explanation:Given: $a^2+b^2+c^2 = ac + \sqrt{3}\,bc$ and $c = 8$.From the Cosine Rule: $a^2 = b^2+c^2 - 2bc\cos A$, $b^2 = a^2+c^2 - 2ac\cos B$, $c^2 = a^2+b^2 - 2ab\cos C$.Adding these three equations gives: $a^2+b^2+c^2 = 2ab\cos C + 2bc\cos A + 2ca\cos B$.Comparing coefficients with the given relation $a^2+b^2+c^2 = ac + \sqrt{3}\,bc$, we obtain:$2\cos C = 0 \;\Rightarrow\; \cos C = 0 \;\Rightarrow\; C = 90^\circ$,$2\cos B = 1 \;\Rightarrow\; \cos B = \frac12 \;\Rightarrow\; B = 60^\circ$,$2\cos A = \sqrt{3} \;\Rightarrow\; \cos A = \frac{\sqrt{3}}{2} \;\Rightarrow\; A = 30^\circ$.Thus the triangle has angles $30^\circ$, $60^\circ$, $90^\circ$; side $c$ is opposite $C=90^\circ$, so $c$ is the hypotenuse.Using sine ratios in the right triangle:$\sin A = \frac{a}{c} \;\Rightarrow\; \sin 30^\circ = \frac{a}{8} \;\Rightarrow\; \frac12 = \frac{a}{8} \;\Rightarrow\; a = 4$,$\sin B = \frac{b}{c} \;\Rightarrow\; \sin 60^\circ = \frac{b}{8} \;\Rightarrow\; \frac{\sqrt{3}}{2} = \frac{b}{8} \;\Rightarrow\; b = 4\sqrt{3}$.Area of the triangle = $\frac12 \times a \times b = \frac12 \times 4 \times 4\sqrt{3} = 8\sqrt{3}$.Answer: $8\sqrt{3}$ (Option C).