Answer
$\theta=\frac{\pi}{2}$
Work Step by Step
Step 1. The area of the triangle can be expressed as $A=\frac{1}{2}ab\ sin\theta$
Step 2. Take the derivatives with respect to the angle to get $A'=\frac{1}{2}ab\ cos\theta$
Step 3. An extreme can be found when $A'=0$, which leads to $\theta=\frac{\pi}{2}$ and $A=\frac{1}{2}ab$
Step 4. This extreme is a maximum because $A''=-\frac{1}{2}ab\ sin\theta \lt 0$ and the function is concave down around $\theta=\frac{\pi}{2}$.
Step 5. We conclude that a maximum area of $A=\frac{1}{2}ab$ happens when $\theta=\frac{\pi}{2}$.
