A jet takes off at an air velocity of 450 km/h with a direction of [E 12° N]. A 15-km/h wind is blowing from the east. Determine the resultant ground velocity. Include a labeled diagram in your solution.

Hello!
The resultant ground velocity of a jet is a vector sum of the jet's air velocity and wind ground velocity. To find its magnitude and direction, we consider the projections on the N-S and W-E axes.
Please look at the attached diagram. The jet's air velocity is in green, the wind velocity is in red and the resultant velocity is in blue. The projections of the air velocity are
450*cos(12^@) approx440.17 (km)/h  and  450*sin(12^@) approx93.56 (km)/h.
The projections of the wind velocity is more obvious, -15 (km)/h and 0.
Hence the resultant velocity has the projections about 425.17 (km)/h and 93.56 (km)/h. The magnitude is equal to
sqrt(425.17^2 +93.56^2) approx435.34 (km)/h.
The direction is tan^(-1)(93.56/425.17) approx 12.41^@.
The answer: the resultant ground velocity is about 435.34 (km)/h in the direction of about E 12.41^@ N.
If the wind velocity is actually 15 m/s, than the idea is absolutely the same, but with the different value (15 m/s =54 (km)/h).