#### Answer

The resulting speed of the boat is 20.85 km/h
The final bearing of the boat is $117.5^{\circ}$

#### Work Step by Step

Let $a = 15~km/h$
Let $b = 7~km/h$
Let angle $\theta$ be the angle between these two vectors. Then $\theta = 130^{\circ}-90^{\circ} = 40^{\circ}$
Let $c$ be the resultant of these two vectors. Note that the magnitude of $c$ is the resulting speed.
We can use the parallelogram rule to find $c$:
$c = \sqrt{a^2+b^2+2ab~cos~\theta}$
$c = \sqrt{(15~km/h)^2+(7~km/h)^2+(2)(15~km/h)(7~km/h)~cos~40^{\circ}}$
$c = \sqrt{434.87~(km/h)^2}$
$c = 20.85~km/h$
The resulting speed of the boat is 20.85 km/h
We can use the law of sines to find the angle A between the vectors b and c:
$\frac{c}{sin~C} = \frac{a}{sin~A}$
$sin~A = \frac{a~sin~C}{c}$
$A = arcsin(\frac{a~sin~C}{c})$
$A = arcsin(\frac{(15)~sin~140^{\circ}}{20.85})$
$A = arcsin(0.46244)$
$A = 27.5^{\circ}$
The final bearing of the boat is $90^{\circ}+27.5^{\circ} = 117.5^{\circ}$