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Question Number 148543 by liberty last updated on 29/Jul/21

Answered by Rasheed.Sindhi last updated on 30/Jul/21

$$Let\mathrm{\angle }\mathrm{B}=\alpha ,\mathrm{\angle }\mathrm{C}=\beta ,$$$$△\mathrm{ABC}:$$$${\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2-2.\mathrm{AB}.\mathrm{BC}.\cos \alpha$$$${\mathrm{AC}}^2=8^2+{10}^2-2.8.10.\cos \alpha$$$$=164-160\cos \alpha$$$$△\mathrm{BCD}:$$$${\mathrm{BD}}^2={10}^2+{12}^2-2.10.12\cos \beta$$$$=244-240\cos \beta$$$$$$$${\mathrm{AC}}^2+{\mathrm{CD}}^2={\mathrm{AD}}^2={\left(2\mathrm{R}\right)}^2$$$${\mathrm{AB}}^2+{\mathrm{BD}}^2={\mathrm{AD}}^2={\left(2\mathrm{R}\right)}^2$$$$Continue$$

Answered by mnjuly1970 last updated on 29/Jul/21

$${AC}^2=164-160cos\left(B\right)$$$${AC}^2=4R^2-144$$$$164-160cos\left(B\right)=4R^2-144$$$$308-4R^2=160cos\left(B\right)$$$$308-4R^2=160(-cos\left(D\right))$$$$4R^2-308=160.\frac{12}{2R}=\frac{960}{R}$$$$R^3-77R-240=0$$$$$$$$$$$$$$$$$$$$$$$$$$

Answered by mr W last updated on 29/Jul/21

$${\sin }^{-1}\frac{4}{R}+{\sin }^{-1}\frac{5}{R}+{\sin }^{-1}\frac{6}{R}=\frac{\pi }{2}$$$$\cos ({\sin }^{-1}\frac{4}{R}+{\sin }^{-1}\frac{5}{R})=\cos (\frac{\pi }{2}-{\sin }^{-1}\frac{6}{R})$$$$\sqrt{(1-\frac{16}{R^2})(1-\frac{25}{R^2})}-\frac{4}{R}\times \frac{5}{R}=\frac{6}{R}$$$$(1-\frac{16}{R^2})(1-\frac{25}{R^2})={(\frac{20}{R^2}+\frac{6}{R})}^2$$$$1=\frac{77}{R^2}+\frac{240}{R^3}$$$$R^3-77R-240=0$$$$R=2\sqrt{\frac{77}{3}}\sin (\frac{2\pi }{3}-\frac{1}{3}{\sin }^{-1}\frac{120\times 3\sqrt{3}}{77\sqrt{77}})$$$$\approx 10.045$$

Commented by liberty last updated on 30/Jul/21

$$\mathrm{great}$$

Answered by mr W last updated on 29/Jul/21

$$anotherway:$$$$AC=\sqrt{4R^2-{12}^2}$$$$BD=\sqrt{4R^2-8^2}$$$$ABCDiscyclic,therefore$$$$AC\times BD=AB\times CD+BC\times AD$$$$\Rightarrow (4R^2-{12}^2)(4R^2-8^2)={(8\times 12+20R)}^2$$$$\Rightarrow R^3-77R-240=0$$$$......$$

Commented by Tawa11 last updated on 03/Aug/21

$$\mathrm{Great}$$

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