[JEE Main 2013] $ABCD$ is a trapezium with $AB\parallel CD$, $BC\perp CD$, $\angle ADB=\theta$, $BC=p$, $CD=q$. Then $AB$ equals
$ABCD$ is a trapezium with $AB\parallel CD$, $BC\perp CD$, $\angle ADB=\theta$, $BC=p$, $CD=q$. Then $AB$ equals
(a) $\dfrac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta}$
(b) $\dfrac{p^2+q^2\cos\theta}{p\cos\theta+q\sin\theta}$
(c) $\dfrac{p^2+q^2}{p^2\cos\theta+q^2\sin\theta}$
(d) $\dfrac{(p^2+q^2)\sin\theta}{(p\cos\theta+q\sin\theta)^2}$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 1mo ago
▲ 0
Correct answer: (a) $\dfrac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta}$
$BD=\sqrt{p^2+q^2}$; in $\triangle ABD$ the sine rule gives $AB=\frac{BD\sin\theta}{\cos(\theta-\phi)}$ where $\tan\phi=\frac qp$, simplifying to $\frac{(p^2+q^2)\sin\theta}{p\cos\theta+q\sin\theta}$.
JEE Main 2013 · Trigonometry — verified solution by the Vidaara Team.
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