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

Answer 1

The correct option is A $\frac{(p^2+q^2)\sin \theta }{p\cos \theta +q\sin \theta }$

Consider $ABCD$ is a trapezium
$\angle ADB=\theta ,BC=p,CD=q$

Let $\angle BDC=\alpha$ then $\angle ABD=\angle BDC=\alpha$
$\angle DAB=\pi -\theta -\alpha$
Applying sine rule on $△ABD$
$\frac{AB}{\sin \theta }=\frac{BD}{\sin (\pi -\theta -\alpha )}$
$AB=\frac{BD\sin \theta }{\sin \theta \cdot \cos \alpha +cos\theta \cdot \sin \alpha }...\left(1\right)$
Now, from $△DCB$
$\sin \alpha =\frac{p}{BD}$, $\cos \alpha =\frac{q}{BD}$
Put the value of $\sin \alpha$ and $\cos \alpha$ in equation $\left(1\right)$
$AB=\frac{BD\sin \theta }{\sin \theta \cdot \frac{q}{BD}+\cos \theta \cdot \frac{p}{BD}}$
$=\frac{B{D}^2\sin \theta }{q\sin \theta +p\sin \theta }$
$=\frac{(p^2+q^2)\sin \theta }{p\sin \theta +q\cos \theta }\because BD=\sqrt{p^2+q^2}$