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  • An are   of a circle subtends a central angle <img alt="\mathrm{\theta}" src="https://entra

An are \mathrm{P Q}  of a circle subtends a central angle \mathrm{\theta}  as shown. Let \mathrm{A(\theta)}  be the area between the chora \mathrm{ P Q}  and the are \mathrm{ P Q}  Let \mathrm{ B(0)}  be the are between the tangent lines \mathrm{ P R}  and \mathrm{ Q R}  and the are \mathrm{ PQ} find  \mathrm{ \operatorname{Lim}_{\theta \rightarrow 0} \frac{A(\theta)}{B(\theta)}} .

Option: 1

1


Option: 2

0


Option: 3

2


Option: 4

4


Answers (1)

best_answer


\mathrm{\begin{aligned} A(\theta)= & \text { Area of sector } P C Q-\text { Area of } \triangle P C R \\ & =\frac{1}{2} r^2 \theta-\frac{1}{2} r^2 \sin \theta=\frac{1}{2} r^2(\theta-\sin \theta) \end{aligned}}

\mathrm{B(\theta)= \text { Arca of quadrilateral } P C Q R-\text { Arca of sector } P C Q R \\

\mathrm{=2 \text { (Area of } \triangle C P R)- \text { area of sector } P C Q P }

\mathrm{=2\left(\frac{1}{2} \cdot r \cdot \tan \frac{\theta}{2}\right)-\frac{1}{2} r^2 \theta=\frac{r^2}{2}\left(2 \tan \frac{\theta}{2}-\theta\right) }

\mathrm{\operatorname{Lim}_{\theta \rightarrow 0} \frac{A(\theta)}{B(\theta)}=-\operatorname{Lim}_{\theta \rightarrow 0} \frac{\theta-\sin \theta}{2 \tan \left(\frac{\theta}{2}\right)-\theta}}

-\lim _{\theta \rightarrow a} \frac{\theta-\left(\theta-\frac{\theta^3}{3 !}+\frac{\theta^4}{4 !}, \ldots .\right)}{2\left(\frac{0}{2}+\frac{\left(\frac{\theta}{2}\right)^3}{3}+\frac{2\left(\frac{\theta}{2}\right)^5}{5 !} \ldots \ldots .\right)-\theta}.

 

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avinash.dongre

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