Answer
The phase angle of the resultant wave is $~~1.55~rad$
Work Step by Step
We can find the horizontal component of the resultant wave's amplitude:
$y_{mh}' = (y_{m1}) cos ~(0) + (y_{m2}) cos~(0.80~\pi)$
$y_{mh}' = (4.60~mm) cos ~(0) + (5.60~mm) cos~(0.80~\pi)$
$y_{mh}' = 0.0695~mm$
We can find the vertical component of the resultant wave's amplitude:
$y_{mv}' = (y_{m1}) sin ~(0) + (y_{m2}) sin~(0.80~\pi)$
$y_{mv}' = (4.60~mm) sin ~(0) + (5.60~mm) sin~(0.80~\pi)$
$y_{mv}' = 3.29~mm$
We can find the phase angle of the resultant wave:
$\beta = tan^{-1}(\frac{3.29~mm}{0.0695~mm})$
$\beta = 1.55~rad$
The phase angle of the resultant wave is $~~1.55~rad$
