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

Answer 1

Answer: (D)

Using sine rule in triangle ABD, we get
$\frac{AB}{\sin \theta }=\frac{BD}{\sin \left(\theta +\beta \right)}\Rightarrow AB=\frac{\sqrt{p^2+q^2}\sin \theta }{\sin \left(\theta +\beta \right)}$

As $\tan \beta =\frac{p}{q}$, we have $\sin \left(\theta +\beta \right)=\sin \theta$
$\cos \beta +\cos \theta \sin \beta$
$=\sin \theta \cdot \frac{q}{\sqrt{p^2+q^2}}+\cos \theta \cdot \frac{p}{\sqrt{p^2+q^2}}$
$=\frac{p\cos \theta +q\sin \theta }{\sqrt{p^2+q^2}}$
We then get $AB=\frac{\left(p^2+q^2\right)\sin \theta }{p\cos \theta +q\sin \theta }$