In a triangle ABC, B=90^{0} and a+b=4. The area of triangle is maximum when C is

  • Option 1)

    \frac{\pi}{4}

  • Option 2)

       \frac{\pi}{3}

  • Option 3)

    \frac{\pi}{6}

  • Option 4)

    None of these

 

Answers (1)

Let angle A is 120-\theta

where c is \theta

Now \frac{\grave{a}}{\sin (120-\theta )}=\frac{b}{\sin 60^o}=K\\ \therefore a=K\sin (120-\theta )=K(\frac{\sqrt{3}}{2}\cos \theta+\frac{1}{2}\sin \theta )

b=K\sin 60^o=\frac{K\sqrt{3}}{2}\\ A=\frac{1}{2}ab\sin \theta ,given\: a+b=4\\ for \:max\:Area \frac{dA}{d\theta }=0\\ \therefore \theta =60^o=\frac{\pi }{3}


Option 1)

\frac{\pi}{4}

This solution is incorrect 

Option 2)

   \frac{\pi}{3}

This solution is correct 

Option 3)

\frac{\pi}{6}

This solution is incorrect 

Option 4)

None of these

This solution is incorrect 

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