Answer
$\frac{\sqrt 6}{3}$
Work Step by Step
$\sin \frac{A}{2}=\pm \sqrt {\frac{1-\cos A}{2}}$
$=\pm\sqrt {\frac{1-(-\frac{1}{3})}{2}}=\pm\sqrt {\frac{2}{3}}=\pm\frac{\sqrt 2}{\sqrt 3}=\pm\frac{\sqrt 6}{3}$
We need to know where $\frac{A}{2}$ terminates to determine the sign of $\sin\frac{A}{2}$.
Given that $90^{\circ}\lt A\lt180^{\circ}$
$\frac{90^{\circ}}{2}\lt\frac{A}{2}\lt\frac{180^{\circ}}{2}$
$45^{\circ}\lt\frac{A}{2}\lt90^{\circ}$
$\frac{A}{2}$ terminates in the first quadrant. In the first quadrant, sine is positive. Therefore,
$\sin\frac{A}{2}=\frac{\sqrt 6}{3}$
