Answer
The phase constant of the resultant wave is $~~45^{\circ}$
Work Step by Step
We can find the horizontal component of the resultant wave's amplitude:
$y_{mh}' = (y_{m1}) cos ~(0) + (y_{m2}) cos~(70^{\circ})$
$y_{mh}' = (3.0~mm) cos ~(0) + (5.0~mm) cos~(70^{\circ})$
$y_{mh}' = 4.71~mm$
We can find the vertical component of the resultant wave's amplitude:
$y_{mv}' = (y_{m1}) sin ~(0) + (y_{m2}) sin~(70^{\circ})$
$y_{mv}' = (3.0~mm) sin ~(0) + (5.0~mm) sin~(70^{\circ})$
$y_{mv}' = 4.70~mm$
We can find the phase constant of the resultant wave:
$tan ~\phi= \frac{4.70~mm}{4.71~mm}$
$\phi= tan^{-1} ~(\frac{4.70~mm}{4.71~mm})$
$\phi = 45^{\circ}$
The phase constant of the resultant wave is $~~45^{\circ}$
