Question

ABCD is a trapezium such that AB is parallel to CD and CB is perpendicular to them.
If $\angle ADB=\theta ,BC=p$ and $CD=q$, show that
$AB=\frac{(p^2+q^2)\sin \theta }{p\cos \theta +q\sin \theta }$

Answer 1

From righanhled $△BCD$, we get
$BD=\sqrt{p^2+q^2}$
Let $\angle ADB=\angle BDC=\alpha$
then $\angle DAB=\pi -(\alpha +\theta )$
and $tan\alpha =p/q.$
Now from $△ABD$, we have
$\frac{AB}{\sin \theta }=\frac{BD}{\sin [\pi -(\theta +\alpha \left)\right]}=\frac{BD}{\sin (\theta +\alpha )}$
$\therefore AB=\frac{BD\sin \theta }{\sin (\theta +\alpha }=\frac{B{D}^2\sin \theta }{BD\sin (\theta +\alpha )}$
$=\frac{B{D}^2\sin \theta }{BD\sin \theta \cos \alpha +BD\cos \theta \sin \alpha }$
=$\frac{(p^2+q^2)\sin \theta }{q\sin \theta +p\cos \theta }$
[$\because B{D}^2=p^2+q^2$ and $BD\cos \alpha =q$ and $BD\sin \alpha =p$.]