Answer
The ship is a distance of 13.5 miles from the port.
The bearing from port is $50.4^{\circ}$
Work Step by Step
Let $a = 10.4~mi$
Let $b = 4.6~mi$
Let angle $C$ be the angle between these two vectors. Then $C = 34^{\circ}+90^{\circ} = 124^{\circ}$
We can use the law of cosines to find $c$, the distance of the ship from the port:
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(10.4~mi)^2+(4.6~mi)^2-(2)(10.4~mi)(4.6~mi)~cos~124^{\circ}}$
$c = \sqrt{182.8~mi^2}$
$c = 13.5~mi$
The ship is a distance of 13.5 miles from the port.
We can use the law of sines to find the angle $B$ between the resultant vector and the vector of $10.4 ~mi$:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{(4.6~mi)~sin~124^{\circ}}{13.5~mi})$
$B = arcsin(0.2825)$
$B = 16.4^{\circ}$
The bearing from port is $34^{\circ}+16.4^{\circ}$ which is $50.4^{\circ}$
