# An isosceles triangle of vertical angle $2\theta$ is inscribed in a circle of radius a. Show the area of the triangle is maximum when $\theta = \frac{\pi }{6}\cdot$

consider the figure shown.

Let $OD= x$
Here $CD= \sqrt{a^{2}-x^{2}}$
Now arc $\left ( ABC \right )= A= \frac{1}{2}\times 2\sqrt{a^{2}-x^{2}} \times \left ( a+x \right )$
Let $s= A^{2}= \left ( a^{2}-x^{2} \right )-\left ( a+x \right )^{2}= \left ( a-x \right )\left ( a+x \right )^{3}$
$\Rightarrow \frac{ds}{dx}= \left ( a-x \right )\times 3\left ( a+x \right )^{2}+\left ( a+x \right )^{3}\left ( -1 \right )$
$\Rightarrow \frac{ds}{dx}= 2\left ( a-2x \right )\left ( a+x \right )^{2}$
$\Rightarrow \frac{d^{2}s}{dx^{2}}= 4\left ( a-2x \right )\left ( a+x \right )-4\left ( a+x \right )^{2}$

for $\frac{ds}{dx}= 2\left ( a-2x \right )\left ( a+x \right )^{2}= 0\; \; \; \therefore x= \frac{9}{2}\; \left [ \because x\neq -a \: \: As \frac{d^{2}s}{dx^{2}} \right ]_{at} x= \frac{a}{2}= -9a^{2}< 0$
$\therefore$ s is maximum at $x= \frac{a}{2}$
$\therefore$ A is also maximum
Now in $\bigtriangleup CDA= \tan \theta = \frac{CD}{AD}= \frac{\sqrt{a^{2}-\frac{a^{2}}{4}}}{a+\frac{a}{2}}= \frac{a\times \frac{\sqrt{3}}{2}}{a\times \frac{\sqrt{3}}{2}}= \frac{1}{\sqrt{3}}$
$\therefore \theta = \frac{\pi }{6}$

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