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Question-141323




Question Number 141323 by mathocean1 last updated on 17/May/21
Answered by mr W last updated on 17/May/21
Commented by mr W last updated on 17/May/21
cos α=((R−r)/(R+r))  BC=((2R)/(tan (α/2)))=2R(√((1+cos α)/(1−cos α)))=2R(√(R/r))
$$\mathrm{cos}\:\alpha=\frac{{R}−{r}}{{R}+{r}} \\ $$$${BC}=\frac{\mathrm{2}{R}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}=\mathrm{2}{R}\sqrt{\frac{\mathrm{1}+\mathrm{cos}\:\alpha}{\mathrm{1}−\mathrm{cos}\:\alpha}}=\mathrm{2}{R}\sqrt{\frac{{R}}{{r}}} \\ $$

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