Trigonometry related problem A trig problem. 

I cannot solve for the distances: $DP_{1,3}$ and $BP_{2,3}$. 
I tried law of sines for the triangle $O_2O_4P_{1,3}$ but don' know the above distances and also don't know another angle. 
Help will be much appreciated.
 A: The image you posted is a bit messy. I will use the picture below and will assume you want to find $e$ and $f$ given $a,b,c,d,\alpha$.

First, draw the line passing through $B$ and $O_4$, I have called it $s$. By the law of cosines,
\begin{equation}
s^2=a^2+b^2-2ab \cos \alpha
\end{equation}
Now you can find $\sin\gamma$ using the law of sines.
\begin{equation}
\sin \gamma= \frac a s \sin \alpha
\end{equation}
Apply the law of cosines again to get $\cos\delta$.
\begin{equation}
\cos\delta= \frac{s^2+d^2-c^2}{2sd}
\end{equation}
Then, 
\begin{equation}
\sin\beta= \sin(\pi-(\pi-\gamma)-\delta)= \sin(\gamma - \delta) = \sin\gamma\cos\delta- \sin\delta\cos\gamma
\end{equation}
Note that $\cos\gamma < 0$. So, the previous equation becomes
\begin{equation}
\sin\beta= \sin\gamma\cos\delta +\sqrt{\bigl(1-\cos^2\delta\bigr)\bigl(1-\sin^2\gamma\bigr)}
\end{equation}
Apply the law of sines to get $e$ and $f$.
\begin{equation}
e= \frac{a \sin \alpha}{\sin\beta} -d \quad \text{and} \quad  f = \frac{s \sin \delta}{\sin\beta}
\end{equation}
P.S. Numerical results are too ugly to be posted... $\color{green} {\ddot\smallsmile}$
