ship leaves port with a bearing of and 27° W. After traveling 16 miles, ship then turned 90° and travels on a bearing of S 63° W 6 miles. At that time, what is the bearing of the ship from port?

Answer 1

The ship has traveled 16 miles, made a left turn (right triangle) then traveled 6 miles. The distance to the port is 6²+16² = d², d= √292 = 17.1. So the ship is 17.1 miles from port.

17.1 miles is the hypotenuse of the triangle so we can figure out the angle.

(r)sin(θ) = 16

(17.1)sin(θ) = 16

cos(θ) = 16/(17.1)

cos^-1(16/17.1) = θ =69.44°

The opposite angle is 90-69.44=20.55°

However that is merely the angle inside the triangle (at the port). We add that to the original 27° to get our bearing

27 + 20.55 = 47.55°

So our bearing is N 47.55° W at 17.1 miles

Answer 2

atan(6/16) counterclockwise from N 27°W = N 27°+20.55° W = N 47.55º W

The 90 degrees in this problem allows us to simplify our work.