Calculate a triangle based on one angle and the lengths on either side of the perpendicular

There is a closed formula for the remaining sides:

$$\begin{align} a&=\sqrt{q^2+h^2}\\ b&=\sqrt{p^2+h^2} \end{align}$$

Where $h$ is calculated using $r$, in the following way:

$$\begin{align} r&=\frac{q+p}{2\sin(\gamma)}\\ h&=\sqrt{r^2-\left(\frac{q-p}{2}\right)^2}\mp\sqrt{r^2-\left(\frac{q+p}{2}\right)^2} \end{align}$$

Where $\gamma\in(0,\pi)$ is the angle at point $C$, and "$\mp$" is resolved as:

$$ \mp=\begin{cases}-,& \text{if }\gamma\ge \pi/2\\+,& \text{if }\gamma\le \pi/2\end{cases} $$

To see how we can obtain this, read the rest of the answer.

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I solved this using the "Inscribed Angle Theorem" mentioned in the comments.

We first observe a triangle with sides $r,q+p,r$ and angle $2\gamma$ above $q+p$.

The Law of Cosines gives fruit (where $q+p=c$):

$$ r=\frac{q+p}{\sqrt{2-2\cos(2\gamma)}}=\frac{c}{2\sin(\gamma)} $$

WLOG the point above $q+p$ is $C'=(0,0)$.

Consider a circle with radius $r$ centered at $C'$, which will be given by $x^2+y^2=r^2$.

To get coordinates of $C$, we use Inscribed Angle Theorem linked by Blue in the comments.

Your point $C=(\frac{q-p}{2},y_c)$ is then the intersection of the circle and the line $x=\frac{q-p}{2}$.

$$ y_c= \pm\sqrt{r^2-\left(\frac{q-p}{2}\right)^2} $$

WLOG assume the positive solution (that $C$ is above $C'$), hence $\pm\to+$.

The height of $B,A$ points is given by Pythagoras' Theorem on the $C'AB$ triangle.

$$ y_b=y_a=\pm\sqrt{r^2-\frac{c^2}{4}} $$

Because of our assumption for $y_c$, here for $y_b=y_a$ the $\pm$ will depend on the $\gamma$. That is, if $2\gamma\ge\pi$ then $C'$ is below $\overline{AB}$, and else if $2\gamma\le\pi$ then $C'$ will be above $\overline{AB}$.

Their $x$-coordinates are simply $x_b=x_c-q$ and $x_a=x_c+p$.

Now we can get the sides $a,b$ as distances $a=d(C,B)$ and $b=d(C,A)$.

$$ a=\sqrt{(x_c-x_b)^2+(y_c-y_b)^2}\\ b=\sqrt{(x_c-x_a)^2+(y_c-y_a)^2} $$

And we have all sides of the triangle now.

An alternative way using the identity

\begin{align} \cot(\gamma_1+\gamma_2) &= \frac{\cot\gamma_1\cot\gamma_2-1}{\cot\gamma_1+\cot\gamma_2} \tag{1}\label{1} . \end{align}

Let $|AB|=c=p+q$, $|CD|=h$, $\angle BCA=\gamma$, $\angle DCA=\gamma_1$, $\angle BCD=\gamma_2$.

\begin{align} \triangle ADC:\quad \cot\gamma_1&=\frac{h}{q} \tag{2}\label{2} ,\\ \triangle BCD:\quad \cot\gamma_2&=\frac{h}{p} \tag{3}\label{3} . \end{align}

Using \eqref{1}, we have

\begin{align} \cot\gamma &= \frac{\cot\gamma_1\cot\gamma_2-1}{\cot\gamma_1+\cot\gamma_2} = \frac{\tfrac{h^2}{pq}-1}{\tfrac hp+\tfrac hq} =\frac{h^2-pq}{(p+q)h} =\frac{h^2-pq}{c\,h} \tag{4}\label{4} . \end{align}

\eqref{4} is equivalent to the quadratic equation in $h$:

\begin{align} h^2-c\,\cot(\gamma)\,h-pq&=0 \tag{5}\label{5} , \end{align}

with two solutions

\begin{align} h_1 &= \tfrac12\,c\,\cot(\gamma) +\tfrac12\,\sqrt{c^2\,\cot^2(\gamma)+4pq} \tag{6}\label{6} ,\\ h_2 &= \tfrac12\,c\,\cot(\gamma) -\tfrac12\,\sqrt{c^2\,\cot^2(\gamma)+4pq} \tag{7}\label{7} . \end{align}

Note that $|c\,\cot(\gamma)|<\sqrt{c^2\,\cot^2(\gamma)+4pq}$, so the expression \eqref{7} would be negative in both cases, whether $\gamma$ is obtuse or acute, that is, whether $\cot\gamma$ is positive or negative, the expression \eqref{7} would be always negative, and the expression \eqref{6} positive, so the only suitable (positive) solution is \begin{align} h &= \tfrac12\,c\,\cot(\gamma) +\tfrac12\,\sqrt{c^2\,\cot^2(\gamma)+4pq} \tag{8}\label{8} . \end{align}

\begin{align} |AB|=c&=p+q ,\\ \end{align}

By the sine rule,

\begin{align} R&=\frac c{2\sin\gamma} . \end{align}

Let $|CE|=h$. Then

\begin{align} S_{ABC}&=\tfrac12ch ,\\ S_{ABC}&=\tfrac12\cdot|AC|\cdot|BC|\sin\gamma \end{align}

$|AC|$ and $|BC|$ in terms of $h$:

\begin{align} |AC|&=\sqrt{q^2+h^2} ,\\ |BC|&=\sqrt{p^2+h^2} , \end{align}

and we have

\begin{align} c^2h^2&=(p^2+h^2)(q^2+h^2)\sin^2\gamma , \end{align}

which results in \begin{align} h^2&= \tfrac12\,(c^2\,\cot^2\gamma+2pq) \pm\sqrt{\tfrac14\,(c^2\,\cot^2\gamma+2pq)^2-p^2q^2} . \end{align}

Edit

Actually, the last expression can be simplified further, and we have a neat single expression for $h$ \begin{align} h&= \tfrac12\,c\,\cot\gamma +\tfrac12\,\sqrt{c^2\cot^2\gamma+4pq} , \end{align} which automatically handles acute/obtuse angle $\gamma$, so the two other side lengths are simply

\begin{align} a&=c\sqrt{\frac pc +\tfrac12\cot\gamma\, \left(\cot\gamma+\sqrt{\cot^2\gamma+4\,\frac {pq}{c^2}} \right)} ,\\ b&=c\sqrt{\frac qc +\tfrac12\cot\gamma\, \left(\cot\gamma+\sqrt{\cot^2\gamma+4\,\frac {pq}{c^2}} \right)} . \end{align}