Question

# $$ABCD$$ 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

A
p2+q2cosθpcosθ+qsinθ
B
p2+q2p2cosθ+q2sinθ
C
(p2+q2)sinθ(pcosθ+qsinθ)2
D
(p2+q2)sinθpcosθ+qsinθ

Solution

## The correct option is D $$\dfrac{(p^{2}+q^{2})\sin\theta}{p\cos\theta+q\sin\theta}$$In the triangle $$BCD$$$$\cos \alpha=\dfrac {q}{\sqrt{p^2+q^2}}$$ and $$\sin \alpha=\dfrac {p}{\sqrt{p^2+q^2}}$$Using sine rule in triangle $$ABD$$$$\dfrac{AB}{\sin\theta}=\dfrac{BD}{\sin(\theta+\alpha)}$$$$\Rightarrow AB=\dfrac{\sqrt{p^{2}+q^{2}}\sin\theta}{\sin\theta\cos\alpha+\cos\theta\sin\alpha}=\dfrac{\sqrt{p^{2}+q^{2}}\sin\theta}{\dfrac{\sin\theta\cdot q}{\sqrt{p^{2}+q^{2}}}+\dfrac{\cos\theta\cdot p}{\sqrt{p^{2}+q^{2}}}}$$$$\Rightarrow AB=\dfrac{(p^{2}+q^{2})\sin\theta}{(p\cos\theta+q\sin\theta)}$$Maths

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