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Question Number 177031 by HeferH last updated on 29/Sep/22

Answered by mr W last updated on 30/Sep/22

Commented by mr W last updated on 30/Sep/22

$${\cos }^{-1}\frac{5}{2R}-{\cos }^{-1}\frac{6}{2R}=30°$$$$\frac{5}{2R}\times \frac{6}{2R}+\sqrt{(1-\frac{25}{4R^2})(1-\frac{36}{4R^2})}=\frac{\sqrt{3}}{2}$$$$(1-\frac{25}{4R^2})(1-\frac{36}{4R^2})={(\frac{\sqrt{3}}{2}-\frac{30}{4R^2})}^2$$$$1-\frac{61}{4R^2}+\frac{25\times 36}{{\left(4R^2\right)}^2}=\frac{3}{4}-\frac{30\sqrt{3}}{4R^2}+\frac{{30}^2}{{\left(4R^2\right)}^2}$$$$\frac{61-30\sqrt{3}}{R^2}=1$$$$\Rightarrow R=\sqrt{61-30\sqrt{3}}$$$$CD=\sqrt{R^2-{\left(\frac{5}{2}\right)}^2}=\sqrt{61-30\sqrt{3}-\frac{25}{4}}=\frac{\sqrt{219-120\sqrt{3}}}{2}$$$$CB=R-r$$$$DE=\frac{r}{\tan 15°}-\frac{5}{2}=(2+\sqrt{3})r-\frac{5}{2}$$$$CD-BE=\frac{\sqrt{219-120\sqrt{3}}}{2}-r$$$${(\frac{\sqrt{219-120\sqrt{3}}}{2}-r)}^2+{((2+\sqrt{3})r-\frac{5}{2})}^2={(\sqrt{61-30\sqrt{3}}-r)}^2$$$$\frac{219-120\sqrt{3}}{4}-\sqrt{219-120\sqrt{3}}r+{(2+\sqrt{3})}^2{r}^2-5(2+\sqrt{3})r+\frac{25}{4}=61-30\sqrt{3}-2\sqrt{61-30\sqrt{3}}r$$$${(2+\sqrt{3})}^{2}r=\sqrt{219-120\sqrt{3}}+5(2+\sqrt{3})-2\sqrt{61-30\sqrt{3}}$$$$\Rightarrow r=\frac{\sqrt{219-120\sqrt{3}}+5(2+\sqrt{3})-2\sqrt{61-30\sqrt{3}}}{{(2+\sqrt{3})}^2}$$$$\approx 1.147828$$

Commented by mr W last updated on 30/Sep/22

$$anotherwaytofindR:$$$$R=\frac{\sqrt{a^2+b^2-2ab\cos \theta }}{2\sin \theta }$$$$=\frac{\sqrt{5^2+6^2-2\times 5\times 6\times \frac{\sqrt{3}}{2}}}{2\times \frac{1}{2}}$$$$=\sqrt{61-30\sqrt{3}}$$

Commented by Tawa11 last updated on 02/Oct/22

$$\mathrm{Great}\mathrm{sir}$$

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