A plane leaves the airport at 4:00 p.m. and heads in the direction N $10 ^ { \circ }$ E at a constant speed of $250$ mi/h. Forty-five minutes later, a jet leaves the same airport and flies N $60 ^ { \circ }$ W at a constant speed of $450$ mi/h. Find the distance between the plane and the jet at 5:30 p.m
A) $409.8 \mathrm { mi }$
B) $904.8 \mathrm { mi }$
C) $309 \mathrm { mi }$
D) $7.5 \mathrm { mi }$
E) $77 \mathrm { mi }$

Question

Quizplus · Accepted Answer

The distance between the plane and the jet can be found using the law of cosines. At 5:30 p.m., the plane has been flying for 1.5 hours and the jet for 1 hour. The distance the plane has flown is $250 \times 1.5 = 375$ miles, and the distance the jet has flown is $450 \times 1 = 450$ miles. The angle between their paths is $70^\circ$ (since $60^\circ + 10^\circ = 70^\circ$). Using the law of cosines, $d = \sqrt{375^2 + 450^2 - 2 \times 375 \times 450 \times \cos(70^\circ)} \approx 409.8$ miles.

A Plane Leaves the Airport at 4:00 P 10∘10 ^ { \circ }10∘

A Plane Leaves the Airport at 4:00 P $10 ^ { \circ }$