Can I find the angle?

Question

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.

Answer 1

Imagine changing $\alpha$ a bit. Now compensate for the change in $a/b$ by moving the top vertex up or down a bit. Now compensate for the change in $D$ by changing $c$ a bit. Now all you've changed is $\alpha$ and $c$, so you can't find $\alpha$ without $c$.

[Edit:]

Perhaps an easier way to see that there can't be a solution is to imagine the line consisting of $a$ and $b$ as a rod that always moves along such that it passes through the corner on the left and the intersection of the vertical line labeled by $h$ with the second oblique line (the one at angle $\theta$). Then if you move the second oblique line and the bottom line upwards together (thus reducing $c$), all that changes is $\alpha$, $a$, $b$ and $c$, whereas $D$, $h$, $x$, $\theta$ and $a/b$ stay the same.

Answer 2

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 $c/D$.

Answer 3

You can compute $x$ and $h$. (BTW, your second equation should be $\cos(\alpha) = (D+x)/a$.). Even then, you cannot determine $\alpha$.

You just got the point $S$. According to the Intercept theorem all the red lines in the figure passing through $S$ have the same ratio $k = a/b$ but they have different angles $\alpha$ to the horizontal line.

EDIT. I am adding the explicit formulae for $x$ and $h$. We have $$ (1) \;\;\; \cos(\alpha) = (D+x) / a$$ $$ (2) \;\;\; \cos(\alpha) = 2D / (a+b)$$

From (1) and (2) we get

$$x = D (k-1) / (k+1) $$

Further, we obtain $h$ as $$ h = x / tan(\theta) $$

Now, the above geometric argument is applied. In addition, we see that $\alpha$ cannot be obtained from the equations (1) and (2) without knowing either $a$ or $a+b$.