Question-164808

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 ° = \frac{\sqrt{3}}{2}

\sin^{2} 15 ° = \frac{2 - \sqrt{3}}{4}

\sin 15 ° = \frac{\sqrt{2 - \sqrt{3}}}{2}

A B = \frac{r}{\sin 15 °}

O B = R - r

O A = R

{(R - r)}^{2} = R^{2} + {(\frac{r}{\sin 15 °})}^{2} - 2 R (\frac{r}{\sin 15 °}) \cos 30 °

\Rightarrow \frac{r}{R} = \frac{\frac{\sqrt{3}}{\sin 15 °} - 2}{\frac{1}{\sin^{2} 15 °} - 1}

= \frac{\frac{2 \sqrt{3}}{\sqrt{2 - \sqrt{3}}} - 2}{\frac{4}{2 - \sqrt{3}} - 1} = \frac{2 (\sqrt{3} - 2 + \sqrt{3 (2 - \sqrt{3})})}{2 + \sqrt{3}}

w i t h R = \frac{1}{2}

\Rightarrow r = \frac{\sqrt{3} - 2 + \sqrt{3 (2 - \sqrt{3})}}{2 + \sqrt{3}} \approx 0.168

Commented by Tawa11 last updated on 22/Jan/22

Great sir .

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