ABCD is a trapezium such that AB and CD are parallel and $\mathrm{BC} \perp \mathrm{CD}$. If $\angle \mathrm{ADB}=\theta, \mathrm{BC}=\mathrm{p}$ and $\mathrm{CD}=\mathrm{q}$, then AB is equal to

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

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

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

(C) $\frac{\left(\mathrm{p}^{2}+\mathrm{q}^{2}\right) \sin \theta}{(\mathrm{p} \cos \theta+\mathrm{q} \sin \theta)^{2}}$

JEE Main · 2013 · 07 Apr 2013 · Offline · Properties of Triangle

ABCD is a trapezium such that AB and CD are parallel and

BC ⊥ CD

. If

∠ ADB = θ, BC = p

and

CD = q

, then AB is equal to

Answer: (D)

In the right triangle

B C D

B D = p^{2} + q^{2}

; let

∠ B D C = α

, so

cos α = \frac{q}{p ^{2} + q ^{2}}

and

sin α = \frac{p}{p ^{2} + q ^{2}}

. Applying the sine rule in triangle

A B D

\frac{A B}{sin θ} = \frac{B D}{sin ( θ + α )}

, and expanding

sin (θ + α) = \frac{q sin θ + p cos θ}{p ^{2} + q ^{2}}

gives

A B = \frac{( p ^{2} + q ^{2} ) sin θ}{p cos θ + q sin θ}

JEE Main · 2013 · 07 Apr 2013 · Offline · Single correct option