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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
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B
p2+q2p2cosθ+q2sinθ
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C
(p2+q2)sinθ(pcosθ+qsinθ)2
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D
(p2+q2)sinθpcosθ+qsinθ
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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 $$ABD$$
$$\dfrac{AB}{\sin\theta}=\dfrac{BD}{\sin(\theta+\alpha)}$$

$$\Rightarrow AB=\dfrac{\sqrt{p^{2}+q^{2}}\sin\theta}{\sin\theta\cos\alpha+\cos\theta\sin\alpha}=\dfrac{\sqrt{p^{2}+q^{2}}\sin\theta}{\dfrac{\sin\theta\cdot q}{\sqrt{p^{2}+q^{2}}}+\dfrac{\cos\theta\cdot p}{\sqrt{p^{2}+q^{2}}}}$$

$$\Rightarrow AB=\dfrac{(p^{2}+q^{2})\sin\theta}{(p\cos\theta+q\sin\theta)}$$

101093_31798_ans.PNG

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