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?

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°