ABCD is trapezium such that AB and CD are parallel and BC perpendicular to CD .If $\overline{)ADB}=\theta$ , BC=p and CD=q then AB=

$\begin{array}{l}\mathrm{△}\mathrm{BCD}\\ {\mathrm{BD}}^2=p^2+q^2\\ BD=\sqrt{p^2+q^2}\\ \text{ Tan }\alpha =\frac{p}{q}\therefore \cos \alpha =\frac{q}{\sqrt{p^2+q^2}};\sin \alpha =\frac{p}{\sqrt{p^2+q^2}}\\ \text{ In }\mathrm{△}A\mathrm{BD}\\ \frac{AB}{\sin \theta }=\frac{BD}{\sin \pi -(\theta +\alpha )}=\frac{BD}{\sin (\theta +\alpha )}\\ AB=\frac{BD\sin \theta }{\sin \theta \cos \alpha +\cos \theta \sin \alpha }\Rightarrow \frac{\sqrt{p^2+q^2}\sin \theta }{\sin \theta \frac{q}{\sqrt{p^2+q^2}}+\cos \theta \frac{p}{\sqrt{p^2+q^2}}}\\ =\frac{\left(p^2+q^2\right)\sin \theta }{p\cos \theta +q\sin \theta }\end{array}$