Question Number 81467 by ajfour last updated on 13/Feb/20
Commented by ajfour last updated on 13/Feb/20
$${If}\:{the}\:{circular}\:{area}\:{cut}\:{out}\:{is} \\ $$$${half}\:{the}\:{sector}\:{area},\:{find}\:\alpha. \\ $$
Answered by ajfour last updated on 13/Feb/20
![(R−r)sin (α/2)=r ((R^2 α)/2)=2πr^2 ⇒ ((R/r))^2 =[((1+sin (α/2))/(sin (α/2)))]^2 =((4π)/α) or (√α)(1+cosec (α/2))=2(√π) ⇒ α=180°, 29.231733°.](https://www.tinkutara.com/question/Q81469.png)
$$\left({R}−{r}\right)\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}={r} \\ $$$$\frac{{R}^{\mathrm{2}} \alpha}{\mathrm{2}}=\mathrm{2}\pi{r}^{\mathrm{2}} \\ $$$$\Rightarrow\:\left(\frac{{R}}{{r}}\right)^{\mathrm{2}} =\left[\frac{\mathrm{1}+\mathrm{sin}\:\left(\alpha/\mathrm{2}\right)}{\mathrm{sin}\:\left(\alpha/\mathrm{2}\right)}\right]^{\mathrm{2}} =\frac{\mathrm{4}\pi}{\alpha} \\ $$$${or}\:\:\:\:\sqrt{\alpha}\left(\mathrm{1}+\mathrm{cosec}\:\frac{\alpha}{\mathrm{2}}\right)=\mathrm{2}\sqrt{\pi} \\ $$$$\Rightarrow\:\alpha=\mathrm{180}°,\:\mathrm{29}.\mathrm{231733}°. \\ $$
Commented by john santu last updated on 13/Feb/20
$$\alpha\:>\:\mathrm{180}^{\mathrm{o}} \:\:?? \\ $$
Commented by ajfour last updated on 13/Feb/20
$${please}\:{look}\:{carefully},\:{Sir}. \\ $$$${there}\:{are}\:{two}\:{values},\:{on}\:{is} \\ $$$$\alpha=\mathrm{180}°. \\ $$