A plane leaves the airport at 400 pm and heads in the direct

A plane leaves the airport at 4:00 p.m. and heads in the direction N E at a constant speed of mi/h. Forty-five minutes later, a jet leaves the same airport and flies N W at a constant speed of mi/h. Find the distance between the plane and the jet at 5:30 p.m.

Solution

distance covered by plane in 45 minutes = 250*.75 = 187.5 miles

distance covered by jet in 45 minutes = 450* .75 = 337.5 miles

the angle between the two planes is 110 degrees

applying law of cosines to find the distance between the two

distance between plane and jet = 438.58 miles