Question Number 25702 by ajfour last updated on 13/Dec/17
Commented by ajfour last updated on 13/Dec/17
$${solution}\:{to}\:{Q}.\:\mathrm{25700} \\ $$
Answered by ajfour last updated on 13/Dec/17
$$\mathrm{2}{x}=\mathrm{1}{cm} \\ $$$${x}+{y}+{z}\:=\mathrm{2}{cm}\:\:\:\:\:\:\:…\left({i}\right) \\ $$$$\mathrm{cot}\:\theta=\frac{{y}+{z}}{{R}}\:\:,\:{and}\:..\left({ii}\right) \\ $$$$\mathrm{sin}\:\theta\:=\frac{{R}−{r}}{{R}+{r}}\:=\frac{{x}}{{x}+{y}+{z}}\:\:\:\:\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow\mathrm{sin}\:\theta=\:\frac{{R}−{r}}{{R}+{r}}=\frac{\mathrm{1}}{\mathrm{4}}\:\:{or}\:\:\:\mathrm{5}{r}=\mathrm{3}{R}\:\:\: \\ $$$$\Rightarrow\:\mathrm{cot}\:\theta\:=\sqrt{\mathrm{15}}\: \\ $$$${considering}\:\left({ii}\right): \\ $$$$\:\:\:\sqrt{\mathrm{15}}\:=\frac{\mathrm{3}}{\mathrm{2}{R}}\:\:\Rightarrow\:{R}=\:\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{15}}}\:=\frac{\sqrt{\mathrm{15}}}{\mathrm{10}}\:. \\ $$$$\:\:\:\:{r}=\frac{\mathrm{3}}{\mathrm{5}}×\frac{\sqrt{\mathrm{15}}}{\mathrm{10}}\:=\frac{\mathrm{3}\sqrt{\mathrm{15}}}{\mathrm{50}}\:. \\ $$