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Q)

ABCD is a trapezium such that AB and CD are parallel and $BC \perp CD$. If $\angle{ADB} = \theta, \; BC = p$ and $CD = q$, then AB is equal to


( A ) $\frac{(p^2+q^2) }{p^2 \cos \theta + q^2 \sin \theta}$
( B ) $\frac{(p^2+q^2) \sin \theta}{p \cos \theta + q \sin \theta}$
( C ) $\frac{(p^2+q^2) \cos \theta}{p \cos \theta + q \sin \theta}$
( D ) $\frac{(p^2+q^2) \sin \theta}{(p \cos \theta + q \sin \theta)^2}$

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