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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6 0

2 years ago

What is negative 1 minus 5?

tatiyna

4 my dudehdhfhfjjeusu

5 0

3 years ago

Read 2 more answers

Two vertical posts stand side by side. One post is 8 feet tall while the other is 17 feet tall. If a 24 foot wire is stretched b

TiliK225 [7]

Answer:24 foot and if i am wrong it is 9 so pick one

Step-by-step explanation:

4 0

3 years ago

What is the exact volume of a sphere with a diameter of 36 in.?

Contact [7]

The volume of a sphere:
$V=\frac{4}{3} \pi r^3$
r - the radius

The diameter is twice the radius.
$d=36 \ in \\
r=\frac{36}{2} \ in = 18 \ in \\ \\
V=\frac{4}{3} \pi \times 18^3=\frac{4}{3}\pi \times 5832=\frac{23328}{3} \pi=7776\pi \ [in^3]$

The exact volume of the sphere is 7776π in³.

5 0

2 years ago