Pretty Simple Physics/Geometry Problem that Stumped Me Okay friends.  So the setup here is that I was playing Silent Hunter III (World War II submarine simulator) and received a radio telling me the position and approximate velocity of a convoy I wanted to raid.  I was trying to calculate the direction I should travel to intercept it most quickly.  To solve this, I set up the problem as a 2D coordinate problem.  I know that's not strictly speaking accurate with curvature of the earth, but I figured it was close enough and it still have me stumped.
So my position is $a_0$, the convoy I want to intercept has position $b_0$ and velocity $\dot b$.  My submarine has a max speed (obviously) and $|\dot a| > | \dot b |$.

To make things a little bit easier on myself, I defined $b_0$ to be $(0,0)$.  Also, to make this easier to conceptualize, I'll give you my actual figures I was working with, but I was trying to solve this algebraically to come up with a general solution.
$$a_0 = (-230, 75)\\ |\dot a| = 30 \\ b_0 = (0,0) \\ |\dot b| = 18.5$$
$b$ was traveling due southwest.
This was the work I managed to figure out:
$$ a(t) = \dot at+a_0 \\ b(t)=\dot bt\\ \text{The ships intersect when their position functions are equal.}\\ \therefore\dot at + a_0 - \dot b t=0\\ \text {Simplifying a bit:}\\(\dot a - \dot b)t+a_0=0$$
So I have two unknowns (i.e. $\dot a$ and $t$) but only one equation.  I'm kind of stuck to figure out my second equation.
Edited to add: I should mention that my units are kilometers for distance and kilometers per hour for speed.
 A: Let $c$ be the point of intersection.
You have a triangle $\triangle abc$ and the law of cosines says $(ab)^2 + (bc)^2-2(ab)(bc)\cos \angle abc = (ac)^2$.
$ab = \sqrt{230^2 + 75^2}$. $ac = |\dot a|t; bc=|\dot b|t$ (where $t$ is the time til intersection; an irrelevant variable which will cancel out) so $\frac {ac}{bc}=\frac {|\dot a|}{|\dot b|}$. And $m\angle abc = 135+\arctan(\frac {75}{230})$
(Actually you said the enemy is going due southwest but your image shows due southeast.  I calculated for simplicity that $m\angle abc = m\angle (-230,75)(0,0)(-230,0) + m\angle(-230,0)(0,0)c$  and $m\angle (-230,75)(0,0)(-230,0)=\arctan \frac {75}{230}$ [ignoring orientation and negative values to keep it simple] and $m\angle (-230,0)(0,0)c = \text{due southeast} = 135$.)
So
$(230^2+75^2) + (bc)^2 - 2\sqrt{230^2 + 75^2}(bc)\cos (135+\arctan\frac {75}{230}) = (bc)^2( \frac {|\dot a|}{|\dot b|})^2$.
That's enough to solve $(bc)$ (there are two solutions but one of them will be a point where $b$ was in the past).  And that's enough to give you $c$.  ANd that gives you $\angle acb$.
