ground speed and direction of plane: 500 mph bearing N35*E

[nickname]

a plane flies at an airspeed (speed still in air) of 500 miles per hour on a bearing of N35°E. An east wind (wind from east to west) is blowing at 30 miles per hour. Find the plane's ground speed and direction.

The answer is 483.4mi/hr, N32.1°E, but how do I solve it.

I did:

x direction: 500sin35+30=316.7882
y direction: 500cos 35= 409.576

ground speed= square root of x^2 +y^2= 517.79mi/hr
direction= 90-tan^-1(y/x)=N37.72degreesE

But the book answers are 483.4mi/hr and N32.1degreesE.

thank you for your help!

 

[BigGlenntheHeavy]

v(sub1) = 500<cos(55),sin(55)> = 500cos(55)i+500sin(55)j

v(sub2) = 30<cos(180),sin(180)> = 30cos(180)i+30sin(180)j

v = v(sub1)+v(sub2) = 256.79i+409.58j

|v| = sqrt(256.79^2+409.58^2) = 483.444 mph (about).

Can you find the angle from here?

 

[nickname]

Yes, thank you very much!