I have worked this problem at least 8 different times and keep getting the same answer every single time. Could somebody please explain how to work this problem?
Here is the problem:
The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45°W at a speed of 30 km/h. (This means that the direction from which the wind blows is 45° west of the northerly direction.) A pilot is steering a plane in the direction N60°E at an airspeed (speed in still air) of 200 km/h. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane. Give your answers correct to one decimal place.
I keep getting 38.6 degrees for the direction and 194.4 for the magnitude.
The way that I worked it:
Resultant:
$$= (-30\sin45 + 200\sin60, 30\sin45 + 200\cos60) $$
$$= (151.9918773, 121.2132034)$$
Direction:
$$= tan^{-1}(\frac{121.2132034}{151.9918773})$$
Magnitude:
$$= \sqrt{121.2132034^2 + 151.9918773^2} $$
Any help would be great.
Thanks!
EDIT: I discovered that the solution to this problem is N67.9$^\circ$E and 209.8 km/h
Best Answer
The resulting vector is:
$$ \vec{v} = \begin{bmatrix} 30\cos(45°) \\ -30\sin(45°)\end{bmatrix} + \begin{bmatrix} 200\cos(30°) \\200\sin(30°)\end{bmatrix} = \begin{bmatrix} 194,42 \\ 78,79 \end{bmatrix} $$
$||\vec{v}||$ = 209,8 m/s under an angle of 22.05 degrees