Question

a plane flies 5.0km due north from a point O and then 6.0km on a bearing of 060° the pilot then changes course on a bearing of 120° for 4.0km. find how far and in what direction the plane is from the starting point?​

Answer 1

The plane flies a distance of approximately 10.536 kilometers in straight line and with a bearing of approximately 035°.

A plane that travels a distance $r$, in kilometers, with a bearing of $\theta$ sexagesimal degrees can be represented in standard position by means of the following expression:

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

We can obtain the resulting vector ($\vec R$) by the principle of superposition:

$\vec R = \Sigma_{i=1}^{n} [r_{i}\cdot (\sin \theta_{i}, \cos \theta_{i})]$ (2)

If we know that $r_{1} = 5\,km$, $\theta_{1} = 0^{\circ}$, $r_{2} = 6\,km$, $\theta_{2} = 60^{\circ}$, $r_{3} = 4\,km$ and $\theta_{3} = 120^{\circ}$, then the resulting vector is:

$\vec R = 5\cdot (\sin 0^{\circ}, \cos 0^{\circ}) + 6\cdot (\sin 60^{\circ}, \cos 60^{\circ}) + 4\cdot (\sin 120^{\circ}, \cos 120^{\circ})$

$\vec R = (5\sqrt{3}, 6) \,[km]$

The magnitude of the resultant is found by Pythagorean theorem:

$\|\vec R\| = \sqrt{R_{x}^{2}+R_{y}^{2}}$

And the bearing is determined by the following inverse trigonometric relationship:

$\theta_{R} = \tan^{-1} \left(\frac{R_{y}}{R_{x}}\right)$ (3)

If we know that $R_{x} = 5\sqrt{3}\,km$ and $R_{y} = 6\,km$, then the magnitude and the bearing of the resultant is:

$\|\vec R\| = \sqrt{(5\sqrt{3})^{2}+6^{2}}$

$\|\vec R\| \approx 10.536\,km$

$\theta_{R} = \tan^{-1} \left(\frac{6}{5\sqrt{3}} \right)$

$\theta_{R} \approx 34.715^{\circ}$

The plane flies a distance of approximately 10.536 kilometers in straight line and with a bearing of approximately 035°.

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