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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
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 & c:d ratios what is the ratio between four regions of the circle when the circle has radius R,the arc divided in a:b has radius r_1 and the arc divided in c:d has radius r_2 .
$$\mathrm{Two}\:\boldsymbol{\mathrm{arcs}}\:\mathrm{having}\:\mathrm{their}\:\mathrm{centers}\:\mathrm{on}\:\mathrm{a} \\ $$$$\boldsymbol{\mathrm{circle}}\:\mathrm{are}\:\mathrm{cutting}\:\mathrm{each}\:\mathrm{other}\:\mathrm{at}\:\mathrm{a}\: \\ $$$$\mathrm{single}\:\mathrm{point}\:\mathrm{inside}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and}\:\mathrm{thus} \\ $$$$\:\mathrm{dividing}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{in}\:\mathrm{four}\:\mathrm{regions}. \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{the}\:\mathrm{arcs}\:\mathrm{cut}\:\mathrm{each}\:\mathrm{other}\:\mathrm{in}\:\boldsymbol{\mathrm{a}}:\boldsymbol{\mathrm{b}}\:\&\:\boldsymbol{\mathrm{c}}:\boldsymbol{\mathrm{d}}\: \\ $$$$\mathrm{ratios}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{between}\:\mathrm{four} \\ $$$$\mathrm{regions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{when}\:\mathrm{the}\:\mathrm{circle} \\ $$$$\mathrm{has}\:\mathrm{radius}\:\boldsymbol{\mathrm{R}},\mathrm{the}\:\mathrm{arc}\:\mathrm{divided}\:\mathrm{in}\:\mathrm{a}:\mathrm{b} \\ $$$$\:\mathrm{has}\:\mathrm{radius}\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{the}\:\mathrm{arc}\:\mathrm{divided}\:\mathrm{in} \\ $$$$\mathrm{c}:\mathrm{d}\:\mathrm{has}\:\mathrm{radius}\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} . \\ $$
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 ((α+β)/2)=(r_1 /(2R)) ⇒α+β=2 cos^(−1) (r_1 /(2R)) (α/β)=(a/b) ⇒α=((2a)/(a+b))× cos^(−1) (r_1 /(2R)) ⇒β=((2b)/(a+b))× cos^(−1) (r_1 /(2R)) ⇒γ+δ=2 cos^(−1) (r_2 /(2R)) ⇒γ=((2c)/(c+d))× cos^(−1) (r_2 /(2R)) ⇒δ=((2d)/(c+d))× cos^(−1) (r_2 /(2R)) let A_1 =area of part IEG ......
$$\mathrm{cos}\:\frac{\alpha+\beta}{\mathrm{2}}=\frac{{r}_{\mathrm{1}} }{\mathrm{2}{R}} \\ $$$$\Rightarrow\alpha+\beta=\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \frac{{r}_{\mathrm{1}} }{\mathrm{2}{R}} \\ $$$$\frac{\alpha}{\beta}=\frac{{a}}{{b}} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}{a}}{{a}+{b}}×\:\mathrm{cos}^{−\mathrm{1}} \frac{{r}_{\mathrm{1}} }{\mathrm{2}{R}} \\ $$$$\Rightarrow\beta=\frac{\mathrm{2}{b}}{{a}+{b}}×\:\mathrm{cos}^{−\mathrm{1}} \frac{{r}_{\mathrm{1}} }{\mathrm{2}{R}} \\ $$$$\Rightarrow\gamma+\delta=\mathrm{2}\:\mathrm{cos}^{−\mathrm{1}} \frac{{r}_{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$$\Rightarrow\gamma=\frac{\mathrm{2}{c}}{{c}+{d}}×\:\mathrm{cos}^{−\mathrm{1}} \frac{{r}_{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$$\Rightarrow\delta=\frac{\mathrm{2}{d}}{{c}+{d}}×\:\mathrm{cos}^{−\mathrm{1}} \frac{{r}_{\mathrm{2}} }{\mathrm{2}{R}} \\ $$$$ \\ $$$${let}\:{A}_{\mathrm{1}} ={area}\:{of}\:{part}\:{IEG} \\ $$$$…… \\ $$

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