Help! I'm Stuck with a Darn Trig Problem!?

Question

Hey… We were given this problem in class and I am having a difficult time solving/ understanding the question. I tried finding the resultant vector and then proceeding from there but then I got all sorts of awkward answers….

Here’s what I tried to find the magnitude:
√((Magnitude)cos (angle) + (Magnitude2) cos(angle2 ) + (Magnitude)sin (angle) + (Magnitude2) sin(angle2 ))

I then used tan to find the angles but this I believe was incorrect. Could someone perhaps attempt the problem and explain/ show how to do it? Thanks a bunch for your help:)

Here’s the problem:

An airplane with an air speed (speed in still air) of 178 m.p.h. is headed due North (90°). If a west wind (180°) of 20m.p.h is blowing, determine the direction travelled by the place and its ground speed (actual speed with respect to the ground).

Answer 1

we need to find the angle that comes about by going 20 to the left and 178 up. say this angle is x

tan(x)=178/20
x=arctan(178/20)=83.589 degrees

the length of the diagonal is the ground speed and we use the equation a^2+b^2=c^2 to find c
sqrt(178^2+20^2)=c=179.12 mph

Answer 2

The problem is somewhat easier if you use standard compass angles, North being 0° and West being 270°. A wind from the West is moving East (90°).
arctanθ = 20/178 ≈ 0.11236
θ ≈ 6.41° "true", or 6.41° E of N
Ground speed
Vg = 20/sin(6.41°) = 178/cos(6.41) ≈ 179.12 mph