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Question Number 164808 by Mathematification last updated on 22/Jan/22

Answered by mr W last updated on 22/Jan/22

Commented by mr W last updated on 22/Jan/22

1−2 sin^2 15°=cos 30°=((√3)/2) sin^2 15°=((2−(√3))/4) sin 15°=((√(2−(√3)))/2) AB=(r/(sin 15°)) OB=R−r OA=R (R−r)^2 =R^2 +((r/(sin 15°)))^2 −2R((r/(sin 15°)))cos 30° ⇒(r/R)=((((√3)/(sin 15°))−2)/((1/(sin^2 15°))−1)) =((((2(√3))/( (√(2−(√3)))))−2)/((4/(2−(√3)))−1))=((2((√3)−2+(√(3(2−(√3))))))/(2+(√3))) with R=(1/2) ⇒r=(((√3)−2+(√(3(2−(√3)))))/(2+(√3)))≈0.168

$$\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{15}°=\mathrm{cos}\:\mathrm{30}°=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\mathrm{sin}^{\mathrm{2}} \:\mathrm{15}°=\frac{\mathrm{2}−\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$$$\mathrm{sin}\:\mathrm{15}°=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$$$ \\ $$$${AB}=\frac{{r}}{\mathrm{sin}\:\mathrm{15}°} \\ $$$${OB}={R}−{r} \\ $$$${OA}={R} \\ $$$$\left({R}−{r}\right)^{\mathrm{2}} ={R}^{\mathrm{2}} +\left(\frac{{r}}{\mathrm{sin}\:\mathrm{15}°}\right)^{\mathrm{2}} −\mathrm{2}{R}\left(\frac{{r}}{\mathrm{sin}\:\mathrm{15}°}\right)\mathrm{cos}\:\mathrm{30}° \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{sin}\:\mathrm{15}°}−\mathrm{2}}{\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \:\mathrm{15}°}−\mathrm{1}} \\ $$$$=\frac{\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}−\mathrm{2}}{\frac{\mathrm{4}}{\mathrm{2}−\sqrt{\mathrm{3}}}−\mathrm{1}}=\frac{\mathrm{2}\left(\sqrt{\mathrm{3}}−\mathrm{2}+\sqrt{\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}\right)}{\mathrm{2}+\sqrt{\mathrm{3}}} \\ $$$${with}\:{R}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\sqrt{\mathrm{3}}−\mathrm{2}+\sqrt{\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}}{\mathrm{2}+\sqrt{\mathrm{3}}}\approx\mathrm{0}.\mathrm{168} \\ $$

Commented by Tawa11 last updated on 22/Jan/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

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