NDA 1 2025 Maths, question 73

Solution.

Answer: (c) $\pi/3$

1. Let $AB = u$ and $AC = 3 - u$. By Pythagoras, $BC = \sqrt{(3 - u)^{2} - u^{2}} = \sqrt{9 - 6u}$, with $0 < u < 3/2$.
2. If the area is $K$, then $4K^{2} = u^{2}(9 - 6u)$.
3. Its derivative is $18u(1 - u)$, positive before $u = 1$ and negative after it.
4. Thus the maximum occurs at $AB = 1$, $AC = 2$.
5. Hence $\cos A = AB/AC = 1/2$, so $A = \pi/3$.