ABCD is a trapezium such that AB and CD are p | vedclass

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Question

(A) Let $AB = x$. Drop a perpendicular from $D$ to $AB$ at point $M$. Then $DM = BC = p$ and $MB = CD = q$. Thus,$AM = AB - MB = x - q$.In $	riangle

Vedclass · Accepted Answer

$\frac{(p^2 + q^2) \sin 	heta}{p \cos 	heta + q \sin 	heta}$

Vedclass · Answer

(A) Let $AB = x$. Drop a perpendicular from $D$ to $AB$ at point $M$. Then $DM = BC = p$ and $MB = CD = q$. Thus,$AM = AB - MB = x - q$.In $	riangle BCD$,$	an \alpha = \frac{p}{q}$.In $	riangle ADM$,$\angle DAM = \pi - (	heta + \alpha)$.Therefore,$	an(\pi - (	heta + \alpha)) = \frac{DM}{AM} = \frac{p}{x - q}$.This implies $-	an(	heta + \alpha) = \frac{p}{x - q}$,or $	an(	heta + \alpha) = \frac{p}{q - x}$.$q - x = p \cot(	heta + \alpha) = p \left( \frac{\cot 	heta \cot \alpha - 1}{\cot \alpha + \cot 	heta} ight)$.Since $\cot \alpha = \frac{q}{p}$,we have $q - x = p \left( \frac{\frac{q}{p} \cot 	heta - 1}{\frac{q}{p} + \cot 	heta} ight) = p \left( \frac{q \cot 	heta - p}{q + p \cot 	heta} ight) = p \left( \frac{q \cos 	heta - p \sin 	heta}{q \sin 	heta + p \cos 	heta} ight)$.$x = q - \frac{pq \cos 	heta - p^2 \sin 	heta}{q \sin 	heta + p \cos 	heta} = \frac{q^2 \sin 	heta + pq \cos 	heta - pq \cos 	heta + p^2 \sin 	heta}{p \cos 	heta + q \sin 	heta}$.$x = \frac{(p^2 + q^2) \sin 	heta}{p \cos 	heta + q \sin 	heta}$.

$ABCD$ 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:

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