Question Number 172437 by infinityaction last updated on 26/Jun/22
Answered by mr W last updated on 27/Jun/22
$${R}={radius}\:{of}\:{semicircle} \\ $$$${r}={radius}\:{of}\:{circle} \\ $$$${O}={center}\:{of}\:{semicircle} \\ $$$${AC}=\mathrm{2}{R}\:\mathrm{cos}\:\theta \\ $$$${AN}={AM}=\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${ON}=\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−{R} \\ $$$$\left(\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−{R}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\frac{{r}}{\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\mathrm{1}\right) \\ $$$$\Rightarrow{r}=\mathrm{2}{R}\left(\mathrm{1}−\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right)\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$${a}=\mathrm{2}{R}\:\mathrm{cos}\:\theta−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\left(\mathrm{cos}\:\theta−\mathrm{1}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$${b}=\mathrm{2}{R}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{cos}\:\theta−\mathrm{1}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\mathrm{sin}\:\theta\:\checkmark \\ $$
Commented by infinityaction last updated on 27/Jun/22
$${thank}\:{you}\:{sir} \\ $$
Commented by Tawa11 last updated on 27/Jun/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$