Aship 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°

The answer to your question is d 68%