If an isosceles triangle of vertical angle $2 \theta$ is inscribed in a circle of radius $a$. Then, area of the triangle is maximum, when $\theta$ is equal to

• Correct answer
  A

  $\frac{\pi}{6}$
• B

  $\frac{\pi}{4}$
• C

  $\frac{\pi}{3}$
• D

  $\frac{\pi}{2}$

Solution

We have, $A D=a+a \cos 2 \theta \text { and } B C=2 B D=2 a \sin 2 \theta$ $\therefore$ Area of $\triangle A B C$ is given by $\Delta=\frac{1}{2} B C \times A D=\frac{2}{2} a^2(\sin 2 \theta+\sin 2 \theta \cos 2 \theta)$ $\Rightarrow \quad \Delta=a^2 \sin 2 \theta+\frac{1}{2} a^2 \sin 4 \theta$
On differentiating both sides w.r.t. $\theta$, we get $\frac{d \Delta}{d \theta}=2 a^2 \cos 2 \theta+2 a^2 \cos 4 \theta$ For maximum or minimum, put $\frac{d \Delta}{d \theta}=0$. $\begin{array}{rlrl}\Rightarrow & \cos 2 \theta & =-\cos 4 \theta \\ \Rightarrow & & 2 \theta & =\pi-4 \theta \\ \Rightarrow & & \theta & =\frac{\pi}{6}\end{array}$ Now, $\frac{d^2 \Delta}{d \theta^2}=-2 a^2 \sin 2 \theta \times 2-2 a^2 \sin 4 \theta \times 4$ $=-4 a^2 \sin 2 \theta-8 a^2 \sin 4 \theta$ At $\quad \theta=\frac{\pi}{6}, \frac{d^2 \Delta}{d \theta^2}=-4 a^2 \sin \frac{\pi}{3}-8 a^2 \sin \frac{2 \pi}{3}$ $\begin{aligned} & =-4 a^2 \times \frac{\sqrt{3}}{2}-8 a^2 \times \frac{\sqrt{3}}{2} \\ & =-6 a^2 \sqrt{3} \lt 0\end{aligned}$ Hence, $\Delta$ is maximum for $\theta=\frac{\pi}{6}$.

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