Question

ABCD is a trapezium such that AB and CD are parallel and CB is perpendicular to then. If $ADB=\theta ,BC=45,CD=30$ and $tan\theta =\frac{1}{5},$ then $51AB$ is equal to

Answer 1

$\frac{AB}{\sin \theta }=\frac{BD}{\sin (\pi -(\theta +\alpha \left)\right)}$
$\Rightarrow AB=\frac{BD\sin \theta }{\sin (\theta +\alpha )}=\frac{BD\sin \theta }{\sin \theta \cos \alpha +\cos \theta \sin \alpha }$
$=\frac{BD\tan \theta }{\sin \alpha +\cos \alpha \tan \theta }$
$=\frac{(BD{)}^2\tan \theta }{45+30\times \frac{1}{5}}[\because \sin \alpha =\frac{45}{BD},\cos \alpha =\frac{30}{BD}]$
$\Rightarrow 51AB=\left[\right(30{)}^2+\left(45{)}^2\right]\times \frac{1}{5}=585$