How to solve this problem with trigonometry and by using vectors? In an assignment I'm asked this:

A boat needs to sail to a port that is on a bearing of 065°. The boat sails in still water at a speed of 35km/h.
  There is a current, moving the water at a speed of 3 km/h on a bearing of 320°.
  What heading should the boat take in order to reach the port?
  You may find it easier to solve this question using trigonometry rather than Cartesian components. You will get the greatest benefit if you try both.

I tried doing the vector part first, where I  found that the boat needs to go at 35.89 km/h in the direction of 69.64 degrees. However, then I realized it's about a sail boat, so you can't choose a speed at random over. I thought first I needed to find the vector for where the boat has to sail to end up going at 35 km/h in the direction of 65 degrees when taking the current into consideration, but now I'm guessing I'm actually supposed to at what angle does a boat sailing at 35 km/h need to be heading to end up going in a direction of 65 degrees after being pushed by a 320-degree 3km/h current.
So I'm a bit stuck as to how I can solve this now. Both in terms of using vector calculations and trigonometry.
 A: I assume that the angles are measured in a clockwise direction starting from a reference direction of $0^\circ$ which is north. 
Using trigonometry the problem can be reduced to find an angle in a triangle knowing two of its sides and another angle.

The water velocity is represented by the blue vector, with a magnitude of $3\, \mathrm{km/h}$. To answer the problem we need to find the direction of the black vector, whose magnitude is $35\, \mathrm{km/h}$. This means we need to find the angle $\theta=A$ of the triangle $\Delta=[A,B,C]$, knowing the sides $a=3$ and $b=35$, and the angle $B=105^\circ$. From the law of sines 
$$\frac{a}{\sin A}=\frac{b}{\sin B}$$
we deduce that
$$\sin\theta=\sin A=\frac{a\sin B}{b}=\frac{3\sin 105^\circ}{35}.$$
So
$$\theta=\arcsin\left( \frac{3\sin 105^\circ}{35}\right)\approx 4.749^\circ .$$
This means that the boat should take the direction of
$$65^\circ+\theta\approx 69.75^\circ,$$
sailing with a speed of $35\, \mathrm{km/h}$. A new application of the law of sines to the side $c$ and angle $C$, yields $c=34.103\, \mathrm{km/h}$, which is the resulting speed of the boat in the port direction.  
A: In the diagram below, the velocity of the boat, $\overrightarrow{BC}$, is in blue, the current, $\overrightarrow{AB}$, is in red and the intended path, $\overrightarrow{AC}$, is in green.
$\hspace{2.5cm}$
Using the Law of Sines, we can get that
$$
\begin{align}
\frac{\sin(\angle BCA)}{3\text{ knots}}
&=\frac{\sin(\angle BAC)}{35\text{ knots}}\\[4pt]
&=\frac{\sin(105^\circ)}{35\text{ knots}}\\[9pt]
\sin(\angle BCA)
&=0.082793642\\[12pt]
\angle BCA
&=4.7491626^\circ
\end{align}
$$
Thus, the heading of the boat should be $69.7491626^\circ$.
A: Hints:


*

*You know the direction of the port (blue line below).

*You know the direction and magnitude of the current (red line)

*You know the magnitude of the boat's speed (green line)

*You are asked to find the direction of the boat.

