$\displaystyle \displaystyle A=\frac{1}{2}bh$
Consider 3 possibilities:
$\displaystyle \displaystyle \frac{\theta}{2}+\frac{\theta}{2}=\theta$
That assume the angle isn't split at first to solve the whole area of the triangle.
Theta =90
Theta < 90
Theta > 90
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 $\displaystyle \displaystyle\theta$.