Question Number 44621 by Tawa1 last updated on 02/Oct/18
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
$$\left.{a}\right){sector}\:{area}\:{s}=\frac{\pi{r}^{\mathrm{2}} }{\mathrm{2}\pi}×\theta=\frac{{r}^{\mathrm{2}} \theta}{\mathrm{2}} \\ $$$$\mathrm{132}=\frac{\mathrm{12}×\mathrm{12}}{\mathrm{2}}×\theta \\ $$$$\theta=\frac{\mathrm{132}×\mathrm{2}}{\mathrm{12}×\mathrm{12}}=\frac{\mathrm{11}}{\mathrm{6}} \\ $$$${l}={r}\theta \\ $$$${l}=\mathrm{12}×\frac{\mathrm{11}}{\mathrm{6}}=\mathrm{22} \\ $$$${so}\:\mathrm{2}\pi{r}_{\mathrm{1}} =\mathrm{22} \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{22}×\mathrm{7}}{\mathrm{2}×\mathrm{22}}=\mathrm{3}.\mathrm{5}{cm} \\ $$
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
$${perpendicula}\:{from}\:{centre}\:{to}\:{chord}\:{pQ}\:{bisect} \\ $$$${PQ} \\ $$$$\mathrm{5}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +{h}^{\mathrm{2}} \:\:\:{h}=\mathrm{4}\:\:\:{distance}\:{from}\:{centre}\:{to}\:{chord}\:{PQ} \\ $$$${let}\:{angle}\:\angle{POQ}=\mathrm{2}\alpha \\ $$$${tan}\alpha=\frac{\mathrm{3}}{\mathrm{4}}\:\:\alpha={tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{37}^{{o}} \\ $$$${so}\:\mathrm{2}\alpha=\mathrm{2}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)=\mathrm{2}×\mathrm{37}^{{o}} =\mathrm{74}^{{o}} \\ $$
Commented by Tawa1 last updated on 02/Oct/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$