So this is the same question as my previous question, A geometry problem (length of side of a quadrilateral), but here we need to find AD 
I need to find AD in terms of AB, BC and $\theta$ only. I tried the same approach as https://math.stackexchange.com/users/1062486/insipidintegrator but did not reach anywhere. It would be help if someone gives some pointers
Following what @ersh said, the exact solutions looks as,
(1) AD = AF + FD = BE + FD
(2) BE = $\frac{BC}{\cos{\theta}}$
(3) $\angle{ADC} = 90 - \theta \because$ Quadrilateral ABCD has two angles 90.
(4) FD = $\frac{EF}{\tan{(\frac{\pi}{2} - \theta})}$ = $\frac{AB}{\tan{(\frac{\pi}{2} - \theta})}$
Therefore,
AD =$\frac{BC}{\cos{\theta}}$ + $\frac{AB}{\tan{(\frac{\pi}{2} - \theta})}$