JEE Main 2013Session, Shift 1MathematicsMediumHeight and Distance38× since 2004ABCDABCDABCD is a trapezium such that ABABAB and CDCDCD are parallel and BC⊥CD.BC \bot CD.BC⊥CD. If ∠ADB=θ, BC=p\angle ADB = \theta ,\,BC = p∠ADB=θ,BC=p and CD=q,CD = q,CD=q, then AB is equal to:AOption A. (p2+q2)sinθpcosθ+qsinθ{{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {p\cos \theta + q\sin \theta }}pcosθ+qsinθ(p2+q2)sinθBOption B. p2+q2cosθpcosθ+qsinθ{{{p^2} + {q^2}\cos \theta } \over {p\cos \theta + q\sin \theta }}pcosθ+qsinθp2+q2cosθCOption C. p2+q2p2cosθ+q2sinθ{{{p^2} + {q^2}} \over {{p^2}\cos \theta + {q^2}\sin \theta }}p2cosθ+q2sinθp2+q2DOption D. (p2+q2)sinθ(pcosθ+qsinθ)2{{\left( {{p^2} + {q^2}} \right)\sin \theta } \over {{{\left( {p\cos \theta + q\sin \theta } \right)}^2}}}(pcosθ+qsinθ)2(p2+q2)sinθCheckSkip · show solution
