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Question:

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

p²+q²/p²cosθ+q²sinθ

(p²+q²)sinθ/(pcosθ+qsinθ)²

p²+q²cosθ/pcosθ+qsinθ

(p²+q²)sinθ/pcosθ+qsinθ

Solution:

In the triangle BCD
cosα = q/√p²+q² and sinα = p/√p²+q²
Using sine rule in triangle ABD
AB/sinθ = BD/sin(θ+α) ⇒ AB = √p²+q² sinθ/sinθcosα + cosθsinα = √p²+q² sinθ/sinθ⋅q/√p²+q² + cosθ⋅p/√p²+q² ⇒ AB = (p²+q²)sinθ/(pcosθ+qsinθ)