Question

Two sides of as triangle are given.Find the angle between them such that the area should be maximum.

Answer 1

Let $a$ and $b$ be the length of given sides and $\theta$ be angle between them. Let $A$ be the area of the triangle.

Then $A=\frac{1}{2}ab\sin \theta \Rightarrow A\left(\theta \right)=\frac{1}{2}ab\sin \theta$

$\therefore$ $A^{{}^{\prime }}\left(\theta \right)=\frac{1}{2}ab\cos \theta A^{{}^{\prime \prime }}\left(\theta \right)=\frac{-1}{2}ab\sin \theta$

$A^{{}^{\prime }}\left(\theta \right)=0\Rightarrow \frac{1}{2}\left(ab\cos \theta \right)=0$

$\Rightarrow \cos \theta =0$

$\theta =\frac{\pi }{2}$

$A^{{}^{\prime \prime }}\left(\frac{\pi }{2}\right)=\frac{-1}{2}ab\sin \frac{\pi }{2}=\frac{-1}{2}ab<0$

Area is maximum at $\theta =\pi /2$.