# An isoceles triangle of vertical angle $2\theta$ is inscribed in a circle of radius a. Then the area of triangle is maximum when $\theta=$? Option 1) $\frac{\pi}{6}$ Option 2) $\frac{\pi}{4}$ Option 3) $\frac{\pi}{3}$ Option 4) $\frac{\pi}{2}$

$\cos 2\theta =\frac{OM}{a}\\ \sin 2\theta =\frac{MC}{a}\\ MC=a\sin 2\theta$

$\therefore A=\frac{1}{2}\times 2a\sin 2\theta \times a(1+\cos 2\theta )\\ a^{2}\sin 2\theta (1+\cos 2\theta )$

$a^{2}(\sin 2\theta +\frac{1}{2}\sin 4\theta )\\ \frac{dA}{d\theta }=a^{2}(2\cos 2\theta +\frac{4}{2}\cos 4\theta )\\ \cos 2\theta +\cos 4\theta =0$for maximum area

$2\cos 3\theta .\cos \theta =0\\ \cos 3\theta =0\\ 3\theta =90^o\\ \theta =30^o\\ \frac{\pi }{6}$

Option 1)

$\frac{\pi}{6}$

This solution is correct

Option 2)

$\frac{\pi}{4}$

This solution is incorrect

Option 3)

$\frac{\pi}{3}$

This solution is incorrect

Option 4)

$\frac{\pi}{2}$

This solution is incorrect

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