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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

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  • $\begingroup$ Just added my work! $\endgroup$ – E. Sawyers Jan 20 '18 at 19:59
  • $\begingroup$ Is this better? $\endgroup$ – E. Sawyers Jan 20 '18 at 20:08
  • $\begingroup$ when the wind blows from the north-west direction, its vector points towards the south-east direction. $\endgroup$ – Kasper Verhulst Jan 20 '18 at 20:13
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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

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  • $\begingroup$ Okay, I have two questions: Why did you use 30deg? Is it because it is 60 degrees from the northern direction, that it is 30deg from the x-axis? The magnitude is correct, but not the direction, why? $\endgroup$ – E. Sawyers Jan 20 '18 at 20:20
  • $\begingroup$ Yea, i took the angle between the plane vector and the x-axis. Quite possibly my calculator returned the resulting angle in radians. Is the correct angle close to 12°? $\endgroup$ – Kasper Verhulst Jan 20 '18 at 20:26
  • $\begingroup$ Hmmm. This might be an issue with web assign. I triple checked your work, and it accepted the magnitude, but not the direction. I keep getting 22.059944846687, so you're correct. It's just not taking the answer as 22.1 or 22.0 $\endgroup$ – E. Sawyers Jan 20 '18 at 20:28
  • $\begingroup$ Hey! I finally figured this out! You were correct about this, but they wanted the true course of the plane. That is basically 90$^\circ$ - 22.1$^\circ$ to get an answer of N67$^\circ$E $\endgroup$ – E. Sawyers Jan 20 '18 at 21:09
  • $\begingroup$ Glad you found it! $\endgroup$ – Kasper Verhulst Jan 21 '18 at 14:46

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