Question

An isosceles triangle of vertical angle $2\theta$ is inscribed in a circle of radius $a$. Then area of the triangle is maximum when $\theta =$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (A)

$\frac{\pi }{6}$

In right $△ODC$,

$OD=acos2\theta$ and $DC=asin2\theta$

So, $AD=a+acos2\theta$

and $BC=2DC=2asin2\theta$

Area of $△ABC=\frac{1}{2}BC.AD$

$A\left(\theta \right)=a^{2}sin2\theta +\frac{a^2}{2}sin4\theta$

$A^{\prime }\left(\theta \right)=2a^{2}cos2\theta +2a^{2}cos4\theta$

For maxima or minima,

$A^{\prime }\left(\theta \right)=0$

$cos2\theta +cos4\theta =0$

$\Rightarrow \theta =\frac{\pi }{2},\frac{\pi }{6}$

$A^{\prime \prime }\left(\theta \right)=-6\sqrt{3}{a}^2$ at $\theta =\frac{\pi }{6}$

Hence, area is maximum at $\theta =\frac{\pi }{6}$