Answer
a) $A=100 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} $
b) $A=50 \sin \theta$ and $\theta =\dfrac{\pi}{2}$
Work Step by Step
a) Let $\dfrac{\theta}{2}$ be half the angle of the triangle with base $b$ and height $h$.
Area, $A=\dfrac{1}{2}\ b \ h...(1)$
We will use the sine and cosine functions to compute base $b$ and height $h$.
Now, $\sin \dfrac{\theta}{2}=\dfrac{b/2}{10} \implies b=20 \sin \dfrac{\theta}{2}$ and $\cos \dfrac{\theta}{2}=\dfrac{h}{10} \implies h=10 \cos \dfrac{\theta}{2}$
Equation (1) becomes: $Area, A=\dfrac{1}{2} (20 \sin \dfrac{\theta}{2}) (10 \cos \dfrac{\theta}{2}) =100 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2} $
b) $Area, A=50 (2 \sin \dfrac{\theta}{2} \cos \dfrac{\theta}{2}) $
$\sin 2x=2 \sin x \cos x$
Now, $A=50 \sin \theta$
To attain a maximum area, $\sin \theta$ must equal $1$. Therefore, $\theta =\dfrac{\pi}{2}$.
