jee-main 2012 Q71

jee-main · India · offline Sine and Cosine Rules Multi-step composite figure problem

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

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

Paper Questions

Q2 Q3 Q4 Q5 Q6 Q12 Q25 Q26 Q30 Q61 Q62 Q63 Q64 Q65 Q66 Q67 Q68 Q69 Q70 Q71 Q72 Q73 Q74 Q75 Q76 Q77 Q78 Q79 Q80 Q81 Q82 Q83 Q84 Q85 Q86 Q87 Q88 Q89 Q90