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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=2R cos θ  AN=AM=(r/(tan (θ/2)))  ON=(r/(tan (θ/2)))−R  ((r/(tan (θ/2)))−R)^2 =(R−r)^2 −r^2   (r/(tan^2  (θ/2)))=2R((1/(tan (θ/2)))−1)  ⇒r=2R(1−tan (θ/2))tan (θ/2)  a=2R cos θ−(r/(tan (θ/2)))=2R(cos θ−1+tan (θ/2))  b=2R−(r/(tan (θ/2)))=2R tan (θ/2)  (a/b)=((cos θ−1+tan (θ/2))/(tan (θ/2)))       =1−((1−cos θ)/(tan (θ/2)))       =1−((2 sin^2  (θ/2))/(tan (θ/2)))       =1−2 sin (θ/2) cos (θ/2)       =1−sin θ ✓

$${R}={radius}\:{of}\:{semicircle} \\ $$$${r}={radius}\:{of}\:{circle} \\ $$$${O}={center}\:{of}\:{semicircle} \\ $$$${AC}=\mathrm{2}{R}\:\mathrm{cos}\:\theta \\ $$$${AN}={AM}=\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${ON}=\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−{R} \\ $$$$\left(\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−{R}\right)^{\mathrm{2}} =\left({R}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$$\frac{{r}}{\mathrm{tan}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}−\mathrm{1}\right) \\ $$$$\Rightarrow{r}=\mathrm{2}{R}\left(\mathrm{1}−\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right)\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$${a}=\mathrm{2}{R}\:\mathrm{cos}\:\theta−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\left(\mathrm{cos}\:\theta−\mathrm{1}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$${b}=\mathrm{2}{R}−\frac{{r}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}=\mathrm{2}{R}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\frac{{a}}{{b}}=\frac{\mathrm{cos}\:\theta−\mathrm{1}+\mathrm{tan}\:\frac{\theta}{\mathrm{2}}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{1}−\mathrm{cos}\:\theta}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\mathrm{2}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\:\mathrm{cos}\:\frac{\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{1}−\mathrm{sin}\:\theta\:\checkmark \\ $$

Commented by infinityaction last updated on 27/Jun/22

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by Tawa11 last updated on 27/Jun/22

Great sir

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

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