In $\triangle A B C, \angle B=90^{\circ}$ and $(b+a)$ is always a constant. In order that $\triangle A B C$ encloses the maximum area, $\angle C=$

• A

  $\frac{\pi}{4}$
• B

  $\frac{\pi}{6}$
• Correct answer
  C

  $\frac{\pi}{3}$
• D

  $\frac{2 \pi}{3}$

Solution

Given, $a+b$ is constant
Let $\quad a+b=k$ Area of $\triangle A B C$ say $A$ $$ \begin{aligned} A & =\frac{1}{2} \times a \times \sqrt{b^2-a^2} \Rightarrow A^2=\frac{1}{4} a^2\left(b^2-a^2\right) \\ A^2 & =\frac{a^2}{4}\left((k-a)^2-c^2\right)=\frac{k^2 a^2-2 k a^3}{4} \end{aligned}$ Differentiating w.r.t. ' $a$ ', $2 A \frac{d A}{d a}=\frac{2 k^2 a-6 k a^2}{4}$ For maximum or minimum $\frac{d A}{d a}=0$ $k^2 a-3 k a^2=0 \Rightarrow a=k / 3 \Rightarrow \frac{d^2 A}{d a^2}=\frac{-k^2}{4 A} < 0$ $A^{\prime}$ s maximum value $a=k / 3$ Now, $a=\frac{k}{3}, b=2 \frac{k}{3} \Rightarrow \cos \theta=\frac{a}{b}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$

Hence, the correct answer is option: 'C'.