How to find phase angle from transfer function
It is just a matter of manipulating complex numbers.

$$ \angle H(\omega) = \tan^{-1} \left( \frac{\Im\{H(\omega)\}}{\Re\{H(\omega)\}} \right)$$

Where \$\Re \{ \cdot \} \$ is the real part and \$ \Im \{ \cdot \} \$ is the imaginary part. (NOTE: this equality is not always strictly true depending on the signs of the real and imaginary parts of \$H(\omega)\$. When finding the angle of an imaginary number the result may need to be adjusted depending on what quadrant the imaginary number is in.)

Expanding \$ H(\omega) \$ gives

$$ H(\omega) = \frac{1}{-\frac{\omega^2}{10}+\frac{11j\omega}{10}+1} $$

Instead of finding the real and imaginary parts of the whole expression, though you could do that, You can note that:

$$ \angle H(\omega) = \angle \text{numerator of } H(\omega) - \angle \text{denominator of } H(\omega) \\
\angle H(\omega) = \tan^{-1}\left(\frac{0}{1}\right) - \tan^{-1}\left(\frac{\frac{\omega}{10}+\omega}{1-\frac{\omega^2}{10}}\right)\\
$$

Using the arctangent addition,wikipedia, formula the expression can be simplified to

$$ \angle H(\omega) = \phi = -\tan^{-1}(\omega)-\tan^{-1}\left(\frac{\omega}{10}\right) $$ 

Basically you get a phase contribution term which is the arctangent of each pole location.