Question

$ABCD$ is a trapezium such that $AB$ and $CD$ are parallel and $BC \bot CD.$ 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 $AB = \frac{\left(p^{2} + q^{2} \right) \sin \theta }{p \cos \theta +q \sin \theta }$