Question

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

Answer 1

A bearing of 34° corresponds to corresponding angle of θ=90-34=56°
The (x,y) values for the position of the ship after completing its first heading are:
x=(10.4cos 56)
y=(10.4sin56)

The trigonometric angle for θ=90-90=0
The (x,y) values for the postion of the ship after completing the second bearing is:
x=(10.4 cos56)+(4.6cos0)≈10.4 mi
y=(10.4sin56)+(4.6sin0)≈8.6mi

the distance from the port will therefore be:
d=√(10.4²+8.6²)≈13.5 miles

It trigonometric angles is:
θ=arctan(y/x)
θ=arctan(8.6/10.4)
θ≈39.6°
Thus the bearing angle is:
90-39.6=50.4°