Question

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Answer 1

The bearing of the plane is approximately 178.037°. $\blacksquare$

Procedure - Determination of the bearing of the plane

Let suppose that bearing angles are in the following standard position, whose vector formula is:

$\vec r = r\cdot (\sin \theta, \cos \theta)$ (1)

Where:

• $r$ - Magnitude of the vector, in miles per hour.
• $\theta$ - Direction of the vector, in degrees.

That is, the line of reference is the $+y$ semiaxis.

The resulting vector ($\vec v$), in miles per hour, is the sum of airspeed of the airplane ($\vec v_{A}$), in miles per hour, and the speed of the wind ($\vec v_{W}$), in miles per hour, that is:

$\vec v = \vec v_{A} + \vec v_{W}$ (2)

If we know that $v_{A} = 239\,\frac{mi}{h}$, $\theta_{A} = 180^{\circ}$, $v_{W} = 10\,\frac{m}{s}$ and $\theta_{W} = 53^{\circ}$, then the resulting vector is:

$\vec v = 239 \cdot (\sin 180^{\circ}, \cos 180^{\circ}) + 10\cdot (\sin 53^{\circ}, \cos 53^{\circ})$

$\vec v = (7.986, -232.981) \,\left[\frac{mi}{h} \right]$

Now we determine the bearing of the plane ($\theta$), in degrees, by the following trigonometric expression:

$\theta = \tan^{-1}\left(\frac{v_{x}}{v_{y}} \right)$ (3)

$\theta = \tan^{-1}\left(-\frac{7.986}{232.981} \right)$

$\theta \approx 178.037^{\circ}$

The bearing of the plane is approximately 178.037°. $\blacksquare$

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