determiner-r-AB-6-AE-5-

Question Number 183631 by a.lgnaoui last updated on 27/Dec/22

d e t e r m i n e r r ?

A B = 6 A E = 5

Commented by a.lgnaoui last updated on 27/Dec/22

Answered by mr W last updated on 28/Dec/22

Commented by mr W last updated on 28/Dec/22

{B E}^{2} = 5^{2} + 6^{2} - 2 \times 5 \times 6 \cos 30 °

\Rightarrow B E = \sqrt{61 - 30 \sqrt{3}}

R = \frac{\sqrt{61 - 30 \sqrt{3}}}{2 \sin 30 °} = \sqrt{61 - 30 \sqrt{3}}

\cos α = \frac{6}{2 \sqrt{61 - 30 \sqrt{3}}} = \frac{3}{\sqrt{61 - 30 \sqrt{3}}}

\Rightarrow \sin α = \frac{\sqrt{52 - 30 \sqrt{3}}}{\sqrt{61 - 30 \sqrt{3}}}

A F = \frac{r}{\sin 15 °} = \frac{4 r}{\sqrt{6} - \sqrt{2}} = (\sqrt{6} + \sqrt{2}) r

\cos (α + 15 °) = \frac{R^{2} + r^{2} {(\sqrt{6} + \sqrt{2})}^{2} - {(R - r)}^{2}}{2 R r (\sqrt{6} + \sqrt{2})}

\cos α \cos 15 ° - \sin α \sin 15 ° = \frac{(7 + 4 \sqrt{3}) r + 2 R}{2 R (\sqrt{6} + \sqrt{2})}

\frac{3}{\sqrt{61 - 30 \sqrt{3}}} \times \frac{\sqrt{6} + \sqrt{2}}{4} - \frac{\sqrt{52 - 30 \sqrt{3}}}{\sqrt{61 - 30 \sqrt{3}}} \times \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{(7 + 4 \sqrt{3}) r + 2 R}{2 R (\sqrt{6} + \sqrt{2})}

3 (2 + \sqrt{3}) - \sqrt{52 - 30 \sqrt{3}} = \frac{(7 + 4 \sqrt{3}) r}{2} + R

\Rightarrow r = 2 (7 - 4 \sqrt{3}) (6 + 3 \sqrt{3} - \sqrt{52 - 30 \sqrt{3}} - \sqrt{61 - 30 \sqrt{3}}) \approx 1.1478

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