Question

an airplane is flying in the direction of 30 degrees south of west at 600 mph. a wind is blowing in the direction of 60 degrees north of east at 40 mph.
are the component forms for the air and wind vectors.
<600cos(30),600sin(30)>;<40cos(60),40sin(60)>
x <600cos(210),600sin(210)>;<40cos(240),40sin(240)>
<600cos(210),600sin(210)>;<40cos(60),40sin(60)>
c=sqrt{(-600\frac{sqrt{3}}{2}+20)^2+(-300 + 40\frac{sqrt{3}}{2})^2}
c=sqrt{(-600\frac{sqrt{3}}{2}-20)^2+(300 - 40\frac{sqrt{3}}{2})^2}

Explanation:

Step1: Determine the angle for the airplane vector

The direction \(30^{\circ}\) South of West. Measured from the positive \(x -\)axis (counter - clockwise), the angle \(\theta = 180^{\circ}+ 30^{\circ}=210^{\circ}\).
For a vector \(\vec{v}\) with magnitude \(|\vec{v}|\) and direction angle \(\theta\), the component form is \(\langle|\vec{v}|\cos\theta,|\vec{v}|\sin\theta
angle\).
For the airplane with \(|\vec{v}_{airplane}| = 600\) mph and \(\theta = 210^{\circ}\), the component form is \(\langle600\cos(210^{\circ}),600\sin(210^{\circ})
angle\).

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Step2: Determine the angle for the wind vector

The direction \(60^{\circ}\) North of East. Measured from the positive \(x -\)axis (counter - clockwise), the angle \(\theta = 60^{\circ}\).
For the wind with \(|\vec{v}_{wind}| = 40\) mph and \(\theta = 60^{\circ}\), the component form is \(\langle40\cos(60^{\circ}),40\sin(60^{\circ})
angle\).

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Answer:

\(\langle600\cos(210^{\circ}),600\sin(210^{\circ})
angle;\langle40\cos(60^{\circ}),40\sin(60^{\circ})
angle\)