Solving angle relationships in triangles with limited information

Question

What is the relationship between angles $\angle abd$ and $\angle acd$, when line $bc=\frac{1}{2}$ and line $ac=30$?

b c and d are in a straight line, and point d is a right angle

The answer should ideally isolate $\angle acd$ from the rest of the equation.

Now, I know this is solvable because if we extend line $ba$ to infinity, and then increase $\angle acd$ from zero up, eventually the line $ac$, which is $30$ long, will hit line $ba$ at a specific angle.

Answer 1

Let's call $\theta=\angle acd$. Then we use a coordinate system with the origin at $b$, and the $x$ axis along $bcd$. Then the coordinates of point $c$ are $(0.5,0)$. The coordinates of point $a$ are $(0.5+30\cos\theta, 30\sin\theta)$. Then $$\tan\angle abd=\frac{30\sin\theta}{0.5+30\cos\theta}$$

Answer 2

Assuming $d$ is a right angle this is a heights/distances problem to find height

$$ ad=h= 30 \sin \angle acd $$

and

$$ bc = h (\cot \angle abd -\cot \angle acd ) $$

Answer 3

Well, thanks to everyone for prodding me on.

I don't know if this is the simplest solution but it works.

Let's add point e onto line $\overline{ab}$ in a position so that a line drawn $\overline{ec}$ is perpendicular to line $\overline{ab}$.

So now:

$\angle acd = \arcsin(\frac{\overline{ad}}{\overline{ac}})$

$\overline{ad} = \sin(\angle abc) * \overline{ab}$

$\overline{ac} = 30$

$\overline{ab} = \overline{ae} + \overline{eb}$

$\overline{ae} = \cos(\angle eac) * \overline{ac}$

$\angle eac = \arcsin(\frac{\overline{ec}}{\overline{ac}})$

$\overline{ec} = \sin(\angle abc) * \overline{bc}$

$\overline{eb} = \cos(\angle abc) * \overline{bc}$

$\overline{bc} = .5$

And all of this expands to:

$$\angle acd = \arcsin(\frac{ \sin(\angle abc) * (\cos(\arcsin(\frac{\sin(\angle abc) * .5}{30})) * 30 + \cos(\angle abc) * .5)}{30})$$