TITLE: Can I find the angle?
QUESTION [6 upvotes]: Look at the diagram

I know $\theta$, $D$ and ratio of $\frac{a}{b}$ (say $k$). I am trying to find out $\alpha$. Is this possible? I couldn't crack it. (I only know the ratio of $a$ to $b$, and I don't know $c$ as well)
UPDATE
I think I have found a solution. Look at this.

x=tan($\theta$)*h
cos($\alpha$)=(2D-x)/a
cos($\alpha$)=2D/(a+b)
ratio of these two => (a+b)/a=2D/(2D-x)
We know a/b and D. We can find x. Since we know $\theta$, we can find $\alpha$.
I am starting a bounty for somebody who can prove this solution wrong. If it is correct, it is mine to keep.
Update2
Ok.. My bad.. I am overall decent guy but fall on my face on this one.

REPLY [2 votes]: I have copied your image below and added some constructions to help find $\tan(\alpha)$.

Note that there are three similar right triangles, each with an angle of $\theta$.  They are similar because they are right triangles and share the opposite (also called vertical) angles near the right arrow under $c\tan(\theta)$.
We can compute the length of the hypotenuse of the right triangle with the dashed, red side and angle of $\theta$ to be $D+c\tan(\theta)$, then compute the dashed, red side to be $D\cos(\theta)+c\sin(\theta)$. From there, it is pretty easy to compute $a$, $a+b$, and then take their ratio.  Solving, we get
$$
\tan(\alpha)=\cot(\theta)-\frac{a/b+1}{a/b}\frac{\cot(\theta)+c/D}{2}
$$
This requires only that we know $a/b$, $c/D$, and $\cot(\theta)$, but $\tan(\alpha)$ is definitely dependent on $c/D$.