I was able to solve this using trig instead of vector math. Here's how it looks as a triangle.
Notice that since this is meant for a computer coordinate system, the
y+ axis is down. Also the angles are as such:
x+ = 0°,
y+ = 90°,
y- = -90°, and
x- = ±180°
Additionally, we know that line B's angle is
∠B = -10°.
The speeds don't matter since they can be expressed as a ratio
r = 120 / 25. So
A can be expressed as
B * r, as shown.
We can calculate the length and angle of
C = sqrt((b.x - a.x)² + (b.y - a.y)²) ≈ 10.05
∠C = atan2(b.y - a.y, b.x - a.x) ≈ 84.29°
∠a = ∠C - ∠B ≈ 94.29°
Now we have all we need to use the Law of Cosine, which states:
A² = B² + C² - 2*B*C*cos(a)
Since we know that
A is a ratio of
B, we can express it like so:
(B*r)² = B² + C² - 2*B*C*cos(a)
B and we get:
B = ± ( sqrt( C² * (cos(a)² + r² -1 ) ) - cos(a)*C ) / (r² - 1) ≈ 2.18
Since time is not a factor in this equation, we get two results, depending on if we're forward in time or backwards. We'd have to check that the
B we get is in the right direction.
Finally, we now know enough to calculate what we want:
c.x = a.x + B * cos(∠B) ≈ 4.14
c.y = a.y + B * sin(∠B) ≈ 1.62
For bonus points, if we wanted the shooting angle
∠A, we can get it like so:
∠A = atan2(c.y - b.y, c.x - b.x) ≈ -83.72°