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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

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

$\frac{p^{2}+q^{2}\cos\theta}{p\cos\theta+q\sin\theta}$

$\frac{p^{2}+q^{2}}{p^{2}\cos\theta+q^{2}\sin\theta}$

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

Option A

Let AB =x

In $\triangle$ DAM, $\tan (\pi-\theta-\alpha)=\frac{p}{x-q}$

$\Rightarrow$ $\tan (\theta+\alpha)=\frac{p}{q-x}$

$\Rightarrow$ $q-x=p\cot(\theta+\alpha)$

$\Rightarrow$ $x=q-p \cot(\theta+\alpha)$

$=q-p\left(\frac{\cot\theta\cot\alpha-1}{\cot\alpha+\cot\theta}\right)$

$\left(\because \cot\alpha=\frac{q}{p}\right)$

= $q-p\left(\frac{\frac{q}{p}\cot\theta-1}{\frac{q}{p}+\cot\theta}\right)$

$=q-p\left(\frac{q\cot\theta-p}{q+p\cot\theta}\right)$

$=q-p\left(\frac{q\cos\theta-p\sin\theta}{q\sin\theta+p\cos\theta}\right)$

$\Rightarrow \frac{q^{2}\sin\theta+pq\cos\theta-pq\cos\theta+p^{2}\sin\theta}{p\cos\theta+q\sin\theta}$

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

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