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This is possibly one of the easiest cal problems ever...

Hi Krystal,
this is not the easiest calculus question ever. To solve this problem, you must figure out (in vector form) both the wind vector (let's call it \[\vec{w}\]and the plane vector \[\vec{P}.\] Then true course of the plane will then be the sum of these two vectors: \[\vec{w}+\vec{P},\] with the ground speed being the norm of this resultant vector: \[\text{ground speed} = || \vec{w} +\vec{P}|| \]
In the picture attached to this message, I have drawn (thanks to geogebra) a planar representation of your situation. Use this key to understand: \[\vec{w}=\overline{AB}, \ \ \ \vec{P}=\overline{AC}\]
the smaller circle is of radius 50 (same as wind speed) and the larger circle is of radius 200 (same as plane speed). To get the coordinates of these two vectors, it's easiest to use polar coordinates. We let due East be 0 degrees, so since the wind vector is on the circle of radius 50, we have:\[\vec{w}=\langle 50\cos(135),50 \sin(135)\rangle=\langle-25\sqrt{2}\,,\ 25\sqrt{2} \rangle.\]
Since the plane vector is on the circle of radius 200, we have \[\vec{P}= \langle 200\cos(60),200\sin(60)\rangle=\langle100\,,\100\sqrt{3} \rangle. \] Hence, the resultant force (or the true course of the plane) is \[\vec{w}+\vec{P}=\langle100-25\sqrt{2}\ ,\ 100\sqrt{3}+25\sqrt{2} \rangle\]
FInally, to find the ground speed you take the norm of this vector:\[||\vec{w}+\vec{P}||=\sqrt{(100-25\sqrt{2})^2 +(100\sqrt{3}+25\sqrt{2})^2}\approx218.349218.\]
The vector coordinates of \[\vec{w}+\vec{P}\] tell you the true course, and the norm of that vector tells you the ground speed.

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