How to calculate a variable vertex's coordinates on a scalene triangle given an original triangle The vertex I'm looking for lies on one of the altitudes of the red triangle which we know everything about via calculation.

Given the desired, final angle (135 degrees, but theoretically, any value), is it possible to calculate the green vertex's coordinates?
I've been told this is an inverse kinematics problem, but maybe it's not necessary to delve into IK yet? Thanks for the help!
 A: Let's label the points.

Construction
Perhaps the easiest solution would be using the inscribed angle theorem to obtain the center of the circle through $A,B,E$. Then you only have to intersect a circle with a line, and choose the right point of intersection.

Since $\angle AEB=130°>90°$, you know that the center of the circle is on the other side of $AB$ than $E$ is. So you consider the supplementary angle, $180°-135°=45°$. The central angle is twice that: $\angle BFA=2\cdot 45°=90°$. And since $\triangle AFB$ is isosceles, the remaining angle sum of $180°-90°$ will divide into equal parts, so $\angle FAB=\angle ABF=45°$. So all you have to do is construct two lines which form a $45°$ angle with the base of the triangle. The point where these intersect is $F$, and the circle around $F$ through $A$ (and $B$) will intersect $CD$ in $E$.
Reproducing the construction
If you want to compute this, there are several possibilities. The most straight-forward approach might be reproducing the construction. You'd could consider difference vectors like $\overrightarrow{AB}=(B-A)$, apply a rotatoion matrix $R$ to that, and use the result as the direction vector of one of the lines. You'd do so for both lines, compute their intersection, and then the radius of the circle. This would lead to a quadratic equation, the solutions of which represent the two points where the line $CD$ intersects the circle. You want the point of intersection which lies within the triangle.
Lengths and the law of sines
You know lengths $AB$, $BC$, $AC$, $AD$ and $BD$ as well as $\angle AEB$ and are looking for length $DE$. The law of sines allows you to compute the following value:
$$d = \frac{\sin\angle AEB}{AB} = \frac{\sin\angle BAE}{BE} = \frac{\frac{DE}{AE}}{BE} = \frac{DE}{AE\cdot BE}$$
You can square this to obtain
$$
d^2=\frac{DE^2}{AE^2\cdot BE^2}=
\frac{DE^2}{(AD^2+DE^2)\cdot(BD^2+DE^2)}
$$
Rename the unknown length $DE$ to $h$ and cross multiply:
\begin{align*}
d^2(AD^2+h^2)(BD^2+h^2)&=h^2
\end{align*}
This is a quadratic equation in $h^2$. Solve that, then find $h$.
Dot product
The dot product (which I write $\langle\cdot,\cdot\rangle$) is proportional to the cosine of the angle:
\begin{align*}
E &= C + \lambda(D-C) = (1-\lambda)C + \lambda D
\\
\langle A-E,B-E\rangle &=
\cos135°\cdot\lVert A-E\rVert\cdot\lVert B-E\rVert
\\
\langle A-E,B-E\rangle^2 &=
\cos^2 135°\cdot\langle A-E,A-E\rangle\cdot\langle B-E,B-E\rangle
\end{align*}
Each of these scalar products can be split into a sum of four, like this:
\begin{multline*}
\langle A-E,B-E\rangle =
(1-\lambda)^2\langle A-C,B-C\rangle \\+
\lambda(1-\lambda)\bigl(\langle A-D,B-C\rangle+\langle A-C,B-D\rangle\bigr) +
\lambda^2\langle A-D,B-D\rangle
\end{multline*}
Do the same for the right hand side of the equation, and you will obtain a quartic equation in $\lambda$.
Degree analysis
I'm a bit surprised by the fact that both of my computational approaches yield fourth-degree conditions, which suggest four distinct solutions might be possible, while on the other hand the geometric approach only allows for two solutions, one of which would result in an incorect angle orientation. I had hoped that perhaps these fourth degrees will somehow simplify, but after actually doing the computation I see that indeed each of these has four distinct algebraic solution. This is probably due to the fact that $\sin$ as well as $\cos$ don't fully describe an angle, but leave a choice of two possibilities. So both approaches will give two extra solutions which correspond to the other angle that has the same $\sin$ or $\cos$.
Hints for the solution
For reference, I'm giving you the polynomial conditions you obtain for angle $135°$. The first approach yields
$$h^4 - 40700h^2 + 30802500 = 0$$
where you can omit the negative solutions, and the solutions where $h>CD$. The other approach results in
$$\lambda^4 - 4\lambda^3 - 16\lambda^2 + 40\lambda - 12 = 0$$
Since $E$ lies between $C$ and $D$, the valid solution also satisfies $0<\lambda<1$.
The coordinates of $E$ for your special case of $135°$ are $(25\sqrt7+m,35\sqrt7-n)$ for suitable $m,n\in\mathbb N$. You can use this to verify the results of your own computations.
