Do these triangles with parallel hypotenuses? This originated as a question that I encountered while woodworking, but nerdy me had to try and see if there was a solution.  Say I have two right triangles, and each of them have the same angles, but one is slightly larger than the other (with the right angle for both at the origin).  Arranging the triangles on top of one another in this way means that the hypotenuses are parallel and separated by a certain amount.  If I pick the most acute angle and name it theta, I know and can express three specific pieces of information (in addition to the obvious piece that one angle is 90 degrees for each triangle):

*

*The length of the opposite side of the larger triangle is 0.5.

*The length of the adjacent side of the smaller triangle is 1.5.

*The separation between the two hypotenuses is 0.125.

The question is, can I solve the remaining pieces of information to fully define these triangles?  I know that you generally need three pieces of info to solve for any given triangle, and I don't have that for each individual triangle, I only have two pieces of information per triangle.  But I do have a fixed and known relationship between the two the separation of the parallel hypotenuses) which I feel like I ought to be able to use in some way as my third piece of information.  For the life of me though, I can't figure out how to do it.  Any ideas?  Am I trying to solve the unsolvable?  Is there a way to do this?  Thanks in advance for the help!
 A: Say the larger triangle is $AOB$ with $A$ on the $+y$-axis, $O$ at the origin and $B$ on the $+x$-axis, and similarly for the smaller triangle $A'OB'$. We have $AO=0.5$ and $OB'=1.5$. Now let $A'O=x$; we then have by similar triangles
$$\frac{1/2-x}{1/8}=\frac{\sqrt{x^2+9/4}}{3/2}$$
This is a quadratic equation. Solving:
$$6-12x=\sqrt{x^2+9/4}$$
$$36-144x+144x^2=x^2+9/4$$
$$143x^2-144x+135/4=0$$
$$x=\frac{144-3\sqrt{159}}{286}=0.3712288\dots\qquad(x<1/2)$$
From there, since we know the triangles are similar, all other data can be computed.
A: In the general case:


$$b\cos\theta = d + a\sin\theta \tag{1}$$

Knowing any three of $a$, $b$, $d$, $\theta$, you can find the fourth. Of course, $\theta$ is the tricky one, but we can write
$$(d+a\sin\theta)^2=b^2\cos^2\theta\quad\to\quad d^2+2a d\sin\theta+a^2\sin^2\theta=b^2(1-\sin^2\theta) \tag{2}$$
Solving the quadratic in $\sin\theta$, we get
$$\sin\theta = \frac{-ad\pm b\sqrt{a^2+b^2-d^2}}{a^2+b^2} \tag{3}$$
For non-negative acute $\theta$ (and non-negative $a$ and $b$), we take the "$\pm$" to be "$+$".
As a woodworker, you would probably prefer to know $\tan\theta$. A little work yields

$$\tan\theta = \frac{ab-d\sqrt{a^2+b^2-d^2}}{(a+d)(a-d)} \tag{4}$$

For the specific case described in the question, we have $a=3/2$, $b=1/2$, $d=1/8$, so
$$\tan\theta = \frac{1}{143} (48 - \sqrt{159}) = 0.2475\ldots \quad\to\quad \theta\approx 13.9^\circ \tag{5}$$
A: 
Let $\angle BEA$ be $\theta$. Then $\angle FDE = \theta$ by similarity, so $\sin \theta = \frac{0.125}{ED} \Rightarrow ED = \frac{0.125}{\sin \theta}$.
Similarly, $\angle BCG = 90º - \theta$, so $\angle CBG = \theta$ as well, and $\cos \theta = \frac{0.125}{BC} \Rightarrow BC = \frac{0.125}{\cos \theta}$.
Since $\Delta ABE \sim \Delta ACD$, we have that:
$$\frac{AB}{AE} = \frac{AC}{AD} \Rightarrow \frac{0.5 - 0.125/\cos \theta}{1.5} = \frac{0.5}{1.5 + 0.125/\sin \theta}$$
$$\Rightarrow \sin \theta \cos \theta (0.5 - 0.125/\cos \theta)(1.5 + 0.125/\sin \theta) = 1.5 \cdot 0.5 \sin \theta \cos \theta$$
$$\Rightarrow \sin \theta \cos \theta (0.75 + 0.0625/\sin \theta - 0.1875/\cos \theta - 0.015625/ (\sin \theta \cos \theta)) = 0.375 \sin \theta \cos \theta$$
$$\Rightarrow 0. 75 \sin \theta \cos \theta +  0.0625 \cos \theta - 0.1875 \sin \theta - 0.015625 = 0.375 \sin \theta \cos \theta$$
$$\Rightarrow \theta \approx 13.9º$$
which gives $AB, AC, AE, AD$, and you can find the hypotenuses $BE$ and $CD$ using Pythagoras.
