Maximize area of isosceles triangle.

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**purplelollipop27** An isosceles triangle has two sides of length (l), and the angle between the sides has measure theta. Determine the angle theta that maximizes the area of the triangle, and find the maximum possible area.

Attachment 19979
I know the answer is pi/2 but i don't know how? Please help!!

Modify your attached figure so that the equal sides have length, $\displaystyle \ell$, rather than 1. Then the altitude, $\displaystyle \displaystyle h$, is given by $\displaystyle \displaystyle h= \ell\cos{\theta\over2}$ and the base, $\displaystyle \displaystyle b$, is given by $\displaystyle \displaystyle b=2\ell\sin{\theta\over2}$

The area is: $\displaystyle \displaystyle A={1\over2}bh={1\over2}\left(2\ell\sin{\theta\over 2}\right)\left(\ell\cos{\theta\over2}\right)$

Use $\displaystyle \displaystyle \sin(2x)=2\sin(x)\cos(x)$, with $\displaystyle \displaystyle x={\theta\over2}$ to simplify this to

$\displaystyle \displaystyle A={1\over2}\,\ell\, ^2\sin\theta$.

Maximize the area with respect to $\displaystyle \displaystyle\theta$.