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Question Number 44621 by Tawa1 last updated on 02/Oct/18

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

$$a)sectorareas=\frac{\pi r^2}{2\pi }\times \theta =\frac{r^2\theta }{2}$$$$132=\frac{12\times 12}{2}\times \theta$$$$\theta =\frac{132\times 2}{12\times 12}=\frac{11}{6}$$$$l=r\theta$$$$l=12\times \frac{11}{6}=22$$$$so2\pi r_1=22$$$$r_1=\frac{22\times 7}{2\times 22}=3.5cm$$

Commented by Tawa1 last updated on 02/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Commented by Tawa1 last updated on 02/Oct/18

$$\mathrm{It}\mathrm{remain}\mathrm{the}\mathrm{b}\mathrm{part}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18

$$perpendiculafromcentretochordpQbisect$$$$PQ$$$$5^2=3^2+h^{2}h=4distancefromcentretochordPQ$$$$letangle\mathrm{\angle }POQ=2\alpha$$$$tan\alpha =\frac{3}{4}\alpha ={tan}^{-1}\left(\frac{3}{4}\right)={37}^o$$$$so2\alpha =2{tan}^{-1}\left(\frac{3}{4}\right)=2\times {37}^o={74}^o$$

Commented by Tawa1 last updated on 02/Oct/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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