Two-arcs-having-their-centers-on-a-circle-are-cutting-each-other-at-a-single-point-inside-the-circle-and-thus-dividing-the-circle-in-four-regions-If-the-arcs-cut-each-other-in-a-b-amp-c-d-ratio

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Question Number 68761 by Rasheed.Sindhi last updated on 15/Sep/19

$$\mathrm{Two}\mathbf{arcs}\mathrm{having}\mathrm{their}\mathrm{centers}\mathrm{on}\mathrm{a}$$$$\mathbf{circle}\mathrm{are}\mathrm{cutting}\mathrm{each}\mathrm{other}\mathrm{at}\mathrm{a}$$$$\mathrm{single}\mathrm{point}\mathrm{inside}\mathrm{the}\mathrm{circle}\mathrm{and}\mathrm{thus}$$$$\mathrm{dividing}\mathrm{the}\mathrm{circle}\mathrm{in}\mathrm{four}\mathrm{regions}.$$$$$$$$\mathrm{If}\mathrm{the}\mathrm{arcs}\mathrm{cut}\mathrm{each}\mathrm{other}\mathrm{in}\mathbf{a}:\mathbf{b}\mathrm{\&}\mathbf{c}:\mathbf{d}$$$$\mathrm{ratios}\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{ratio}\mathrm{between}\mathrm{four}$$$$\mathrm{regions}\mathrm{of}\mathrm{the}\mathrm{circle}\mathrm{when}\mathrm{the}\mathrm{circle}$$$$\mathrm{has}\mathrm{radius}\mathbf{R},\mathrm{the}\mathrm{arc}\mathrm{divided}\mathrm{in}\mathrm{a}:\mathrm{b}$$$$\mathrm{has}\mathrm{radius}{\mathbf{r}}_1\mathrm{and}\mathrm{the}\mathrm{arc}\mathrm{divided}\mathrm{in}$$$$\mathrm{c}:\mathrm{d}\mathrm{has}\mathrm{radius}{\mathbf{r}}_2.$$

Commented by Rasheed.Sindhi last updated on 15/Sep/19

Answered by mr W last updated on 18/Sep/19

Commented by mr W last updated on 18/Sep/19

$$\cos \frac{\alpha +\beta }{2}=\frac{r_1}{2R}$$$$\Rightarrow \alpha +\beta =2{\cos }^{-1}\frac{r_1}{2R}$$$$\frac{\alpha }{\beta }=\frac{a}{b}$$$$\Rightarrow \alpha =\frac{2a}{a+b}\times {\cos }^{-1}\frac{r_1}{2R}$$$$\Rightarrow \beta =\frac{2b}{a+b}\times {\cos }^{-1}\frac{r_1}{2R}$$$$\Rightarrow \gamma +\delta =2{\cos }^{-1}\frac{r_2}{2R}$$$$\Rightarrow \gamma =\frac{2c}{c+d}\times {\cos }^{-1}\frac{r_2}{2R}$$$$\Rightarrow \delta =\frac{2d}{c+d}\times {\cos }^{-1}\frac{r_2}{2R}$$$$$$$$let{A}_1=areaofpartIEG$$$$\dots \dots$$

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