Question Number 165514 by Tawa11 last updated on 02/Feb/22
Answered by som(math1967) last updated on 03/Feb/22
$$\left.\boldsymbol{{a}}\right)\:\:\boldsymbol{{let}}\:\boldsymbol{{radius}}\:\boldsymbol{{OC}}=\boldsymbol{{x}} \\ $$$$\therefore\boldsymbol{{OD}}=\left(\boldsymbol{{x}}−\mathrm{2}\right) \\ $$$$\boldsymbol{{from}}\bigtriangleup\boldsymbol{{OED}} \\ $$$$\:\:\boldsymbol{{x}}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} +\left(\boldsymbol{{x}}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\boldsymbol{{x}}^{\mathrm{2}} =\mathrm{36}+\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{4}\boldsymbol{{x}}+\mathrm{4}\Rightarrow\boldsymbol{{x}}=\mathrm{10} \\ $$$$\left.\boldsymbol{{b}}\right)\: \\ $$$$\angle\boldsymbol{{EOC}}=\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{6}}{\mathrm{10}}=\mathrm{37}^{°} °\left(\boldsymbol{{aprox}}\right) \\ $$$$\boldsymbol{{area}}\:\boldsymbol{{of}}\:\boldsymbol{{sector}} \\ $$$$=\frac{\mathrm{37}}{\mathrm{360}}×\boldsymbol{\pi}×\mathrm{10}^{\mathrm{2}} \boldsymbol{{sqm}} \\ $$
Commented by som(math1967) last updated on 03/Feb/22
Commented by Tawa11 last updated on 03/Feb/22
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$