Answer
The liner is a distance of 6.6 miles from the port.
The bearing from port is $117.1^{\circ}$
Work Step by Step
Let $a = 8.8~mi$
Let $b = 2.4~mi$
Let angle $C$ be the angle between these two vectors. Then $C = 90^{\circ}-70^{\circ} = 20^{\circ}$
We can use the law of cosines to find $c$, the distance of the liner from the port:
$c = \sqrt{a^2+b^2-2ab~cos~C}$
$c = \sqrt{(8.8~mi)^2+(2.4~mi)^2-(2)(8.8~mi)(2.4~mi)~cos~20^{\circ}}$
$c = \sqrt{43.5~mi^2}$
$c = 6.6~mi$
The liner is a distance of 6.6 miles from the port.
We can use the law of sines to find the angle $B$ between the resultant vector and the vector of $8.8 ~mi$:
$\frac{c}{sin~C} = \frac{b}{sin~B}$
$sin~B = \frac{b~sin~C}{c}$
$B = arcsin(\frac{b~sin~C}{c})$
$B = arcsin(\frac{(2.4~mi)~sin~20^{\circ}}{6.6~mi})$
$B = arcsin(0.12437)$
$B = 7.1^{\circ}$
The bearing from port is $110^{\circ}+7.1^{\circ}$ which is $117.1^{\circ}$