As usual, draw a diagram. To find the distance z, it's a simple law of cosines problem.
z^2 = 92^2 + 50^2 - 2*92*cos117°
z = 123.048
Getting the bearing is a bit more work. You have to use x- and y-components of the vectors so you can add them up. Starting at (0,0), you have
<0,-90>+<50cos(-153°),50sin(-153°)>
= <-44.55,-112.7>
so, using the standard angles,
tanθ = -44.55/-112.7
θ = -112°
As a bearing from due north, that is 90-θ:
202° or S22°W
A plane flies due south from an airport for 92 miles, and then turns and goes S63°W for 50
miles. How far is the plane from the airport (to the nearest tenth of a mile)? What direction is the plane
from the airport (to the nearest tenth of a degree)? (That is, find the bearing from the airport to the plane.)
1 answer
1 answer