Question

Consider $ABCD$ is a trapezium such that $AB,DC$ are parallel and $BC$ is perpendicular to them. If $\angle ADB=\theta ,BC=p$ and $CD=q,$ then $AB$ is equal to

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

Answer 1

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


in the above figure, $\angle C={90}^{\circ }$
Using sine rule,
$\frac{AB}{\sin \theta }=\frac{BD}{\sin (\theta +\alpha )}\therefore \frac{AB}{\sin \theta }=\frac{\sqrt{p^2+q^2}}{\sin \theta \cos \alpha +\cos \theta \sin \alpha }\cdots \left(i\right)$
But in $△BCD,\sin \alpha =\frac{p}{\sqrt{p^2+q^2}},\cos \alpha =\frac{q}{\sqrt{p^2+q^2}}$
putting these values in $\left(i\right),$
$\therefore AB=\frac{(p^2+q^2)\sin \theta }{q\sin \theta +p\cos \theta }$