So you have two vectors 125,40° and 60,165°
Let's first find the x and y components...
x=125cos40+60cos165 and y=125sin40+60sin165
The magnitude will found by using the pythagorean theorem:
v=√(x^2+y^2), using x and y found about you will get:
v≈103.06
And the direction is found using the identity tanα=y/x so
α=arctan(y/x), which again give x and y found earlier:
α≈68.483° in the second quadrant (because x is negative) so
α≈180-68.483≈111.52°
So the resultant vector is approximately:
103.06, 111.52°
