Problem: In a triangle $ABC$, right angled at $B$, $AP$ is the median. $Q$ is a point taken on side $AB$ such that $\angle ACQ =\alpha $ and $\angle APQ =\beta $ and let $CP = BP = x$, Find $AQ$ in terms of the given parameters.
What I have tried is draw perpendiculars from $Q$ on $AC$ and on $AP$ to get some trigonometry going but I am unable to proceed so I graphed it on the coordinate plane but I'm unable to proceed there too. So I looked at the options (its a multiple choice) and all of them seem to include $\cot \alpha$ and $\cot \beta$.
Rough Figure:
Edit: Thanks for the help everyone. Also, I am sorry to learn that I forgot to mention that I had assumed the distance $|AB|$ and it isn't given in the question itself.
Let $|AC|=b,\ |AP|=u,\ |CQ|=v,\ |PQ|=w$.
\begin{align} b^2&=4\,x^2+c^2 ,\quad v^2=4\,x^2+(c-q)^2 ,\\ u^2&=\phantom{4\,}x^2+c^2 ,\quad w^2=\phantom{4\,}x^2+(c-q)^2 ,\\ b^2-u^2&= v^2-w^2=3\,x^2 ,\\ b^2-v^2&=u^2-w^2 =2\,c\,q-q^2 . \end{align}
Considering squared areas of $\triangle CAQ$ $\triangle PAQ$, which share the same base $|AQ|=q$ and $h_{CAQ}=2h_{PAQ}$, we have
\begin{align} v^2\,b^2\,\sin^2\phi &= 4\,u^2\,w^2\,\sin^2\psi \tag{1}\label{1} . \end{align}
\begin{align} \triangle CAQ:\quad 2\,x\,q&= b\,v\,\sin\phi \tag{2}\label{2} ,\\ \triangle PAQ:\quad \phantom{2\,}x\,q&= u\,w\,\sin\psi \tag{3}\label{3} ,\\ u\,w&= b\,v\,\frac{\sin\phi}{2\,\sin\psi} \tag{4}\label{4} . \end{align}
By cosine rule
\begin{align} \triangle CAQ:\quad q^2&= b^2+v^2-2\,b\,v\,\cos\phi \tag{5}\label{5} ,\\ \triangle CAQ:\quad q^2&= u^2+w^2-2\,u\,w\,\cos\psi \tag{6a}\label{6a} \\ &=u^2+w^2-b\,v\,\sin\phi\,\cot\psi \tag{6b}\label{6b} ,\\ b\,v&= \frac{b^2+v^2-u^2-w^2}{2\,\cos\phi-\sin\phi\cot\psi} \tag{7}\label{7} ,\\ b\,v&= \frac{6\,x^2}{2\,\cos\phi-\sin\phi\cot\psi} \tag{8}\label{8} ,\\ u\,w&= \frac{3\,x^2}{2\,\sin\psi\,\cot\phi-\cos\psi} \tag{9}\label{9} . \end{align}
From $\triangle PAQ$
\begin{align} 2\,x\,q&= b\,v\sin\phi \tag{10}\label{10} ,\\ q&=\frac{b\,v\sin\phi}{2\,x} = \frac{3\,x\,\sin\phi}{2\,\cos\phi-\sin\phi\cot\psi} = \frac{3\,x}{2\,\cot\phi-\cot\psi} \tag{11}\label{11} . \end{align}
\begin{align} b^2+v^2&= q^2+2\,b\,v\,\cos\phi \tag{12}\label{12} ,\\ b^2+v^2&= q^2 +\frac{12\,x^2}{2\cos\phi-\sin\phi\cot\psi} \tag{13}\label{13} \end{align}
Substitution of \eqref{1} gives a quadratic equation in $c$:
\begin{align} c^2-q\,c+4\,x^2-\frac{6\,x^2\,\cos\phi}{2\,\cos\phi-\sin\phi\cot\psi} &=0 \tag{14}\label{14} ,\\ c^2-q\,c+\tfrac23\,q\,x\,(\cot\phi-2\,\cot\psi)&=0 \tag{15}\label{15} \end{align}
with a suitable root
\begin{align} c&=\frac{q}{2}\,\left(1 +\sqrt{1-\frac{8\,x}{3\,q}\,(\cot\phi-2\,\cot\psi)} \right) \tag{16}\label{16} ,\\ c&=\frac x2\cdot\frac{3+\sqrt{9+8\,\cot\phi\cot\psi-16\,(\cot\phi-\cot\psi)^2}} {2\,\cot\phi-\cot\psi} \tag{17}\label{17} . \end{align}