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Question Number 125444 by ajfour last updated on 11/Dec/20

Commented by ajfour last updated on 11/Dec/20

$$Findradiusofthequartercircle.$$

Answered by ajfour last updated on 11/Dec/20

Commented by ajfour last updated on 11/Dec/20

$$\mathrm{\angle }BAT=30°$$$$AT=\sqrt{3}$$$$AC=\sqrt{R^2+1}=x$$$$\sin \theta =\frac{1}{AC}=\frac{1}{x}$$$$T\equiv (1-AT\sin \theta ,AT\cos \theta )$$$$\equiv (1-\frac{\sqrt{3}}{x},\frac{\sqrt{3}\sqrt{x^2-1}}{x})$$$$B\equiv (x_T+\cos \theta ,y_T+\sin \theta )$$$$\equiv (1-\frac{\sqrt{3}}{x}+\frac{\sqrt{x^2-1}}{x},\frac{\sqrt{3}\sqrt{x^2-1}}{x}+\frac{1}{x})$$$$letcenterofquartercirclebe$$$$originO.$$$$OB=R-1=\sqrt{x^2-1}-1$$$${(1-\frac{\sqrt{3}}{x}+\frac{\sqrt{x^2-1}}{x})}^2+{(\frac{\sqrt{3}\sqrt{x^2-1}}{x}+\frac{1}{x})}^2$$$$={(\sqrt{x^2-1}-1)}^2$$$$\Rightarrow$$$${(x-\sqrt{3}+\sqrt{x^2-1})}^2+{(\sqrt{3}\sqrt{x^2-1}+1)}^2$$$$=x^2(x^2-2\sqrt{x^2-1})$$$$\Rightarrow$$$$2x^2-1+3-2\sqrt{3}x+2(x-\sqrt{3})\sqrt{x^2-1}$$$$+3(x^2-1)+1+2\sqrt{3}\sqrt{x^2-1}$$$$=x^4-2x^2\sqrt{x^2-1}$$$$\Rightarrow$$$$x^4-5x^2+2\sqrt{3}x=2x(x+1)\sqrt{x^2-1}$$$$\Rightarrow (x^3-5x+2\sqrt{3})=2(x+1)\sqrt{x^2-1}$$$$\Rightarrow$$$$(x-\sqrt{3})(x^2+x\sqrt{3}-2)=2(x+1)\sqrt{x^2-1}$$$$Isthereawaytosolveitexactly?$$$$calculatorgives$$$$x\approx 3.58238505$$$$R=\sqrt{x^2-1}\approx 3.43998294$$

Commented by ajfour last updated on 11/Dec/20

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