#### Answer

3

#### Work Step by Step

x is in quadrant II, where
its sine is positive, and cosine is negative.
Furthermore,
$90^{o} < \theta < 180^{o}\qquad/\div 2$
$45^{o} < \displaystyle \frac{\theta}{2} < 90^{o}$,
so $\displaystyle \frac{\theta}{2}$ is in quadrant I, where $\displaystyle \tan\frac{\theta}{2}$ is positive.
The Half-Angle Identity:
$\displaystyle \tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$
.. has the plus sign, we need $\cos\theta$ ...
Pythagorean identity, ($\cos x$ is negative):
$\cos\theta=-\sqrt{1-\sin^{2}\theta}=-\sqrt{1-\frac{9}{25}}$
$=-\displaystyle \sqrt{\frac{16}{25}}=-\frac{4}{5}=-0.8$
So,
$\displaystyle \tan\frac{x}{2}=+\sqrt{\frac{1-(-0.8)}{1+(-0.8)}}=\sqrt{\frac{1.8}{0.2}}$
$=\sqrt{9}=3$