Question
An airplane travels due west for 2.5 hours at 340 miles per hour. Then it changes course to S51°W. Find the airplanes distance from its point of departure and its bearing, after a total flight time of 4.5 hours.
Answers
Steve
Using the law of cosines, the distance z is
z^2 = (2.5*340)^2 + (2*340)^2 - 2(2.5*340)(2*340)cos(141°)
z = 1443
bearing: 90-θ where
tanθ = (2*340*cos51°)/(2.5*340+2*340*sin51°)
θ = E 17.25° N
or 72.75°
z^2 = (2.5*340)^2 + (2*340)^2 - 2(2.5*340)(2*340)cos(141°)
z = 1443
bearing: 90-θ where
tanθ = (2*340*cos51°)/(2.5*340+2*340*sin51°)
θ = E 17.25° N
or 72.75°