Solve the problem.
-An airplane travels at $105 \mathrm {~km} / \mathrm { h }$ for $5 \mathrm { hr }$ in a direction of $321 ^ { \circ }$ from Greenville. At the end of this time, how far west of Greenville is the plane (to the nearest kilometer)?
A) $330 \mathrm {~km}$
B) $648 \mathrm {~km}$
C) $408 \mathrm {~km}$
D) $834 \mathrm {~km}$

Question

Quizplus · Accepted Answer

To find how far west the plane is, we need to calculate the westward component of the plane's displacement. The plane's total displacement is $105 \mathrm{~km/h} \times 5 \mathrm{~hr} = 525 \mathrm{~km}$. The direction of $321^\circ$ is measured counterclockwise from the positive x-axis (east direction), which means the westward component is equivalent to the sine of the angle from the north (since $321^\circ$ is $39^\circ$ west of north). Thus, the westward displacement is $525 \sin(39^\circ) \approx 330 \mathrm{~km}$, to the nearest kilometer.

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