Question

$ABCD$ 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 ) \, \sin \, \theta}{ p \, \cos \, \theta + q \, \sin \, \theta}$
• B. $ \frac{ p^2 + q^2 \, \cos \, \theta}{ p \, \cos \, \theta + q \, \sin \, \theta}$
• C. $ \frac{ p^2 + q^2}{ p^2 \, \cos \, \theta + q^2 \, \sin \, \theta}$
• D. $ \frac{ (p^2 + q^2) \, \sin \, \theta}{ (p \, \cos \, \theta + q \, \sin \, \theta)^2}$

Answer 1

Answer: (A)

$B D=\sqrt{p^{2}+q^{2}}$
$\angle A B P=\angle B D C=\alpha$
$\Rightarrow \angle D A B=\pi-(\theta+\alpha)$
$\tan \alpha=\frac{p}{q}$
$\triangle A B D$
$\frac{A B}{\sin \theta} =\frac{B D}{\sin (\pi-(\theta+\alpha))}$
$=\frac{B D}{\sin (\theta+\alpha)} $
$\therefore A B= \frac{B D \sin \theta}{\sin (\theta+\alpha)}$
$=\frac{B D^{2} \sin \theta}{B D \sin (\theta+\alpha)}=\frac{B D^{2} \sin \theta}{B D \sin \theta \cos x+B D \cos \theta \cos x}$
$=\frac{\left(P^{2}+\cos ^{2}\right) \sin \theta}{q \sin \theta +P \cos \theta} $