A ship leaves port on a bearing of 340 degrees and travels 1

A ship leaves port on a bearing of 34.0 degrees and travels 10.4 mi. The ship then turns due east and travels 4.6 mi. How far is the ship from port, and what is its bearing from the port?

Solution

The (x, y) coordinate after travelling 10.4 mi at bearing of 34 deg

=(10.4*sin34°,10.4*cos34°) = (5.82,8.62)

Now go E for 4.6, and that adds (4.6,0) to the displacement. Now you are at

(10.42,8.62)

So, Distance of ship from port = sqrt(10.42^2 + 8.62^2) = 13.52 miles

The new bearing is given by = tan^-1(8.62/10.42) = 39.60 deg

Bearing = 90 - 39.60 = 50.4 deg

tan = 10.42/8.62 = 1.2088
= 50.4°
distance is sqrt(182.88) = 13.52