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

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

In the triangle $BCD$

$\cos \alpha =\frac{q}{\sqrt{p^2+q^2}}$ and $\sin \alpha =\frac{p}{\sqrt{p^2+q^2}}$

Using sine rule in triangle $ABD$

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

$\Rightarrow AB=\frac{\sqrt{p^2+q^2}\sin \theta }{\sin \theta \cos \alpha +\cos \theta \sin \alpha }=\frac{\sqrt{p^2+q^2}\sin \theta }{\frac{\sin \theta \cdot q}{\sqrt{p^2+q^2}}+\frac{\cos \theta \cdot p}{\sqrt{p^2+q^2}}}$

$\Rightarrow AB=\frac{(p^2+q^2)\sin \theta }{(p\cos \theta +q\sin \theta )}$