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}

 

Answers (1)
V Vakul

\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 =0for 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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