સમલંબ ચતુષ્કોણ $ABCD$ માટે, $\overline{AB}\parallel\overline{CD}$ તેમજ $\overline{BC}\bot\overline{CD}$ આપેલ હોવાથી, આકૃતિમાં દર્શાવ્યા મુજબ કાટકોણ $\triangle BDC$ રચાશે. જ્યાં $m\angle BDC=\alpha$
$\therefore BD^2=BC^2+CD^2=p^2+q^2 .......................(1)$
અને $\cos\alpha=\frac{CD}{BD}=\frac{q}{\sqrt{p^2+q^2}}......... ........(2)$ તેમજ $\sin\alpha=\frac{BC}{BD}=\frac{p}{\sqrt{p^2+q^2}}..........................(3)$
વળી, $\triangle ABD$ માટે, $\sin $ સુત્રનો ઉપયોગ કરતા,
$\frac{AB}{\sin\theta}=\frac{BD}{\sin A}$ મળે.
$\therefore \frac{AB}{\sin\theta}=\frac{\sqrt{p^2+q^2}}{\sin(\pi-(\theta+\alpha))} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ((1)$ પરથી)
$\frac{AB}{\sin\theta}= \frac{\sqrt{p^2+q^2}}{\sin\theta\cos\alpha+\cos\theta\sin\alpha}$
$\therefore AB= \frac{\sqrt{p^2+q^2}\sin\theta}{\sin\theta\left(\frac{q}{\sqrt{p^2+q^2}}\right)+\cos\theta \left(\frac{p}{\sqrt{p^2+q^2}}\right)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ((2),(3)$ પરથી
$\therefore AB = \frac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.