Calculate the angle and long side of a parallelogram that is defined by a right triangle Ok, so I need to calculate the angle and the length of long sides of a parallelogram that is defined my a right triangle. The top line of the parallelogram intersects the top corner (a) of the right triangle and the bottom line of the parallelogram intersects the bottom corner (b) of the right triangle. I also know the height of the parallelogram.
How can I accomplish this? I took Trig about 10 years ago and I've lost about 90% of it.
See image for details blue is what I need to know, black is what I do know.
(I did this in SolidWorks if you are wondering).

 A: Label the left side, bottom and hypotenuse of the triangle $a$, $b$ and $c$, respectively, and the unknown angle $\theta$. In addition, let the length of the short side of the parallelogram be $d$. We have the following relationships among the various lengths: $$a=c\cos\theta \\ b=c\sin\theta=12\frac78 \\d\sin\theta=1\frac12 \\ a+d=11\frac14$$ You thus have four equations in four unknowns, so it should be possible to work through them to find $c$ and $\theta$. To begin with, we have from the second and third equations $\frac23d=\frac8{103}c$, so one of those variables can be eliminated right away. Substituting for $a$ and $d$ in the last equation produces $$c\cos\theta+{12\over103}c=\frac{45}4.$$ From the second equation we have $\sin\theta={103\over8c}$, so the preceding equation becomes $$c\sqrt{1-\left({103\over8c}\right)^2}+{12\over103}c=\frac{45}4.$$ Solving this for $c$ is ugly, but straightforward. From the second equation at the top, $\theta=\arcsin{103\over8c}$, which you can look up or plug into a calculator once you have $c$.
A: We have the following two equations:
$$ 11.25 = \frac{1.5}{ sin( \color{blue}{53.89^{\circ}} ) }+ \color{red}{15.9375} \cdot cos( \color{blue}{53.89^{\circ}} ) $$
$$ (11.25 - \frac{1.5}{sin( \color{blue}{53.89^{\circ}} ) })^2 + 12.875^2 = \color{red}{15.9375}^2 $$
