Answer
$-\frac{\sqrt {15}}{3}$
Work Step by Step
Step 1. Using the given figures, we have $a=-\sqrt {5-4}=-1, b=-\sqrt {1-\frac{1}{16}}=-\frac{\sqrt {15}}{4}$
Step 2. $h(\frac{\alpha}{2})=tan(\frac{\alpha}{2})=\frac{1-cos\alpha}{sin\alpha}=\frac{1-(-1/4)}{-\frac{\sqrt {15}}{4}}=-\frac{\sqrt {15}}{3}$