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Question-205808

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Question Number 205808 by cortano12 last updated on 31/Mar/24
Commented by A5T last updated on 02/Apr/24
I guess this is not uniquely defined.
$${I}\:{guess}\:{this}\:{is}\:{not}\:{uniquely}\:{defined}.\: \\ $$
Commented by A5T last updated on 02/Apr/24
Yea, that′s true,makes it unique.
$${Yea},\:{that}'{s}\:{true},{makes}\:{it}\:{unique}. \\ $$
Commented by mr W last updated on 02/Apr/24
AD should tangent the semi−circle.
$${AD}\:{should}\:{tangent}\:{the}\:{semi}−{circle}. \\ $$
Answered by mr W last updated on 31/Mar/24
Commented by mr W last updated on 31/Mar/24
tan (α/2)=(R/(2x)) sin α=(x/R) (x/R)=((2×(R/(2x)))/(1+((R/(2x)))^2 )) say t=(x/R) t=((1/t)/(1+(1/(4t^2 ))))=((4t)/(4t^2 +1)) ⇒t=((√3)/2) tan θ=((2x)/(2R))=(x/R)=t=((√3)/2) ⇒sin θ=((√3)/( (√(2^2 +((√3))^2 ))))=(√(3/7))
$$\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}=\frac{{R}}{\mathrm{2}{x}} \\ $$$$\mathrm{sin}\:\alpha=\frac{{x}}{{R}} \\ $$$$\frac{{x}}{{R}}=\frac{\mathrm{2}×\frac{{R}}{\mathrm{2}{x}}}{\mathrm{1}+\left(\frac{{R}}{\mathrm{2}{x}}\right)^{\mathrm{2}} } \\ $$$${say}\:{t}=\frac{{x}}{{R}} \\ $$$${t}=\frac{\frac{\mathrm{1}}{{t}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }}=\frac{\mathrm{4}{t}}{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{t}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{x}}{\mathrm{2}{R}}=\frac{{x}}{{R}}={t}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }}=\sqrt{\frac{\mathrm{3}}{\mathrm{7}}} \\ $$
Answered by A5T last updated on 02/Apr/24
AD=AC=2x; Let AD and CE meet at F Then, ((FE)/(FC=FE+CE))=((DE)/(AC))=(1/2)⇒2FE=FE+CE ⇒FE=CE; ((AD)/(AF))=(1/2)⇒AF=4x ⇒FC=(√(AF^2 −AC^2 ))=(√(16x^2 −4x^2 ))=2x(√3)⇒CE=x(√3) (x(√3))(2r−x(√3))=x^2 ⇒(2r=(x/( (√3)))+x(√3))=((4x)/( (√3))) sinθ=((AC=2x)/( (√(((16x^2 )/3)+4x^2 ))))=((2x)/(x(√((28)/3))))=((2(√3))/( (√(28))=2(√7)))=((√3)/( (√7)))
$${AD}={AC}=\mathrm{2}{x};\:{Let}\:{AD}\:{and}\:{CE}\:{meet}\:{at}\:{F} \\ $$$${Then},\:\frac{{FE}}{{FC}={FE}+{CE}}=\frac{{DE}}{{AC}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{2}{FE}={FE}+{CE} \\ $$$$\Rightarrow{FE}={CE};\:\frac{{AD}}{{AF}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{AF}=\mathrm{4}{x} \\ $$$$\Rightarrow{FC}=\sqrt{{AF}^{\mathrm{2}} −{AC}^{\mathrm{2}} }=\sqrt{\mathrm{16}{x}^{\mathrm{2}} −\mathrm{4}{x}^{\mathrm{2}} }=\mathrm{2}{x}\sqrt{\mathrm{3}}\Rightarrow{CE}={x}\sqrt{\mathrm{3}} \\ $$$$\left({x}\sqrt{\mathrm{3}}\right)\left(\mathrm{2}{r}−{x}\sqrt{\mathrm{3}}\right)={x}^{\mathrm{2}} \Rightarrow\left(\mathrm{2}{r}=\frac{{x}}{\:\sqrt{\mathrm{3}}}+{x}\sqrt{\mathrm{3}}\right)=\frac{\mathrm{4}{x}}{\:\sqrt{\mathrm{3}}} \\ $$$${sin}\theta=\frac{{AC}=\mathrm{2}{x}}{\:\sqrt{\frac{\mathrm{16}{x}^{\mathrm{2}} }{\mathrm{3}}+\mathrm{4}{x}^{\mathrm{2}} }}=\frac{\mathrm{2}{x}}{{x}\sqrt{\frac{\mathrm{28}}{\mathrm{3}}}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{28}}=\mathrm{2}\sqrt{\mathrm{7}}}=\frac{\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{7}}} \\ $$
Commented by A5T last updated on 02/Apr/24

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