Question

ABCD is a trapezium such that $AB\parallel CD$and CB is perpendicular to them.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. $\frac{(p^2+q^2)\cos \theta }{p\sin \theta +q\cos \theta }$
• C. $\frac{(p^2+q^2)\sin \theta }{p\sin \theta +q\cos \theta }$
• D. $\frac{(p^2+q^2)\cos \theta }{p\cos \theta +q\sin \theta }$

Answer 1

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

Let $\angle ABD=\angle BDC=\alpha$ [alternate angles]

$\therefore \angle BAD={180}^{\circ }-(\theta +\alpha )$

By the sine formula,in $\Delta ABD$ we have

$\frac{AB}{\sin \theta }=\frac{BD}{\sin ({180}^{\circ }-(\theta +\alpha \left)\right)}$

$\Rightarrow AB=\frac{BD\sin \theta }{\sin (\theta +\alpha )}=\frac{BD\sin \theta }{\sin \theta \cos \alpha +\cos \theta \sin \alpha }$

In $\Delta BCD,\sin \alpha =\frac{p}{BD}$ and $\cos \alpha =\frac{q}{BD}$

Also $B{D}^2=p^2+q^2$

Therefore,from Eq.(i) we have $AB=\frac{BD\sin \theta }{\sin \theta \left(\frac{q}{BD}\right)+\cos \theta \left(\frac{p}{BD}\right)}=\frac{B{D}^2\sin \theta }{q\sin \theta +p\cos \theta }=\frac{(p^2+q^2)\sin \theta }{p\cos \theta +q\sin \theta }$