Question: $\mathrm{ABCD}$ is a trapezium such that $\mathrm{AB}$ and $\mathrm{CD}$ are parallel and $\mathrm{BC} \perp \mathrm{CD}$. If $\angle \mathrm{ADB}=\theta, \mathrm{BC}=$ $p$ and $C D=q$, then $A B$ is equal to
$\frac{\left(p^{2}+q^{2}\right) \sin \theta}{p \cos \theta+q \sin \theta}$
$\frac{p^{2}+q^{2} \cos \theta}{p \cos \theta+q \sin \theta}$
$\frac{p^{2}+q^{2}}{p^{2} \cos \theta+q^{2} \sin \theta}$
$\frac{\left(p^{2}+q^{2}\right) \sin \theta}{(p \cos \theta+q \sin \theta)^{2}}$
Correct Option: 1,
Solution:
