Discussion Forum



1)

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

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

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

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

Answer:

Option A

Explanation:

Let AB =x

  542021744_m6.JPG

 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)$

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

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

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

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

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

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

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