The bearing of the plane is approximately 178.037°. [tex]\blacksquare[/tex]
Let suppose that bearing angles are in the following standard position, whose vector formula is:
[tex]\vec r = r\cdot (\sin \theta, \cos \theta)[/tex] (1)
Where:
That is, the line of reference is the [tex]+y[/tex] semiaxis.
The resulting vector ([tex]\vec v[/tex]), in miles per hour, is the sum of airspeed of the airplane ([tex]\vec v_{A}[/tex]), in miles per hour, and the speed of the wind ([tex]\vec v_{W}[/tex]), in miles per hour, that is:
[tex]\vec v = \vec v_{A} + \vec v_{W}[/tex] (2)
If we know that [tex]v_{A} = 239\,\frac{mi}{h}[/tex], [tex]\theta_{A} = 180^{\circ}[/tex], [tex]v_{W} = 10\,\frac{m}{s}[/tex] and [tex]\theta_{W} = 53^{\circ}[/tex], then the resulting vector is:
[tex]\vec v = 239 \cdot (\sin 180^{\circ}, \cos 180^{\circ}) + 10\cdot (\sin 53^{\circ}, \cos 53^{\circ})[/tex]
[tex]\vec v = (7.986, -232.981) \,\left[\frac{mi}{h} \right][/tex]
Now we determine the bearing of the plane ([tex]\theta[/tex]), in degrees, by the following trigonometric expression:
[tex]\theta = \tan^{-1}\left(\frac{v_{x}}{v_{y}} \right)[/tex] (3)
[tex]\theta = \tan^{-1}\left(-\frac{7.986}{232.981} \right)[/tex]
[tex]\theta \approx 178.037^{\circ}[/tex]
The bearing of the plane is approximately 178.037°. [tex]\blacksquare[/tex]
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