ABCD is a trapezium such that AB and CD are parallel and BC perp CD, if angle ADB = θ, BC = p and CD = q, then AB is equal to

Question

ABCD is a trapezium such that AB and CD are parallel and BC perp CD, if angle ADB = θ, BC = p and CD = q, then AB is equal to (A)  ( (p2 + q2 ) sin θ/ p cos θ + q sin θ) (B)  ( p2 + q2 cos θ/ p cos θ + q sin θ) (C)  ( p2 + q2/ p2 cos θ + q2 sin θ) (D)  ( (p2 + q2) sin θ/ (p cos θ + q sin θ)2)

Tardigrade Admin · Accepted Answer

B D=√p2+q2 angle A B P= angle B D C=α ⇒ angle D A B=π-(θ+α) tan α=(p/q) triangle A B D (A B/ sin θ) =(B D/ sin (π-(θ+α))) =(B D/ sin (θ+α)) ∴ A B= (B D sin θ/ sin (θ+α)) =(B D2 sin θ/B D sin (θ+α))=(B D2 sin θ/B D sin θ cos x+B D cos θ cos x) =((P2+ cos 2) sin θ/q sin θ +P cos θ)

Q. $A BC D$ is a trapezium such that $A B$ and $C D$ are parallel and $BC$ $⊥$ $C D$ , if $∠$ $A D B = θ, BC = p$ and $C D = q$ , then $A B$ is equal to

$\frac{( p ^{2} + q ^{2} ) s i n θ}{p c o s θ + q s i n θ}$

$\frac{p ^{2} + q ^{2} c o s θ}{p c o s θ + q s i n θ}$

$\frac{p ^{2} + q ^{2}}{p ^{2} c o s θ + q ^{2} s i n θ}$

$\frac{( p ^{2} + q ^{2} ) s i n θ}{( p c o s θ + q s i n θ ) ^{2}}$

Q. 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

pcosθ+qsinθ(p2+q2)sinθ​

pcosθ+qsinθp2+q2cosθ​

p2cosθ+q2sinθp2+q2​

(pcosθ+qsinθ)2(p2+q2)sinθ​

BD=p2+q2​ ∠ABP=∠BDC=α ⇒∠DAB=π−(θ+α) tanα=qp​ △ABD sinθAB​=sin(π−(θ+α))BD​ =sin(θ+α)BD​ ∴AB=sin(θ+α)BDsinθ​ =BDsin(θ+α)BD2sinθ​=BDsinθcosx+BDcosθcosxBD2sinθ​ =qsinθ+Pcosθ(P2+cos2)sinθ​

Q. $A BC D$ is a trapezium such that $A B$ and $C D$ are parallel and $BC$ $⊥$ $C D$ , if $∠$ $A D B = θ, BC = p$ and $C D = q$ , then $A B$ is equal to

$\frac{( p ^{2} + q ^{2} ) s i n θ}{p c o s θ + q s i n θ}$

$\frac{p ^{2} + q ^{2} c o s θ}{p c o s θ + q s i n θ}$

$\frac{p ^{2} + q ^{2}}{p ^{2} c o s θ + q ^{2} s i n θ}$

$\frac{( p ^{2} + q ^{2} ) s i n θ}{( p c o s θ + q s i n θ ) ^{2}}$