What is $∠ A$ equal to if the area of the triangle is maximum?

Question

Question

Consider the following for the two (02) items that follow: Let ABC be a triangle right-angled at B and AB+AC = 3 units.

This question was previously asked in

NDA-I (Mathematics) Official Paper (Held On: 13 Apr, 2025)

Answer 1

Calculation:

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Given,

$AB + AC = 3$

Let $AB = x$ and $AC = 3 - x$.

Then,

$BC = \sqrt{AC^2 - AB^2} = \sqrt{(3 - x)^2 - x^2} = \sqrt{9 - 6x} $

The area of the triangle is,

$A = \tfrac12\,x\,\sqrt{9 - 6x} $

To maximize, set up

$A^2 = \tfrac14\,x^2\,(9 - 6x)$

$\displaystyle \frac{d(A^2)}{dx} = \tfrac14\bigl(2x(9 - 6x) + x^2(-6)\bigr) = \frac{18x(1 - x)}{4} = 0 $

Hence $x = 1$ (discarding x=0, so

$BC = \sqrt{9 - 6}= \sqrt{3},\quad AC = 3 - 1 = 2$

Therefore,

$\sin A = \frac{BC}{AC} = \frac{\sqrt{3}}{2} \implies A = \frac{\pi}{3}$

∴ $\angle A = \frac{\pi}{3}$.

Hence, the correct answer is Option 3.

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