Question Number 208581 by Tawa11 last updated on 18/Jun/24
Commented by mr W last updated on 18/Jun/24
Commented by Tawa11 last updated on 18/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Answered by A5T last updated on 18/Jun/24
Commented by A5T last updated on 18/Jun/24
$$\mathrm{2}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}×\mathrm{5}{cos}\theta=+\mathrm{2}×\mathrm{3}×\mathrm{4}{cos}\theta \\ $$$$\Rightarrow{cos}\theta=\frac{\mathrm{1}}{\mathrm{11}} \\ $$$${AB}=\sqrt{\mathrm{2}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}×\mathrm{2}×\mathrm{5}×\frac{\mathrm{1}}{\mathrm{11}}}=\frac{\sqrt{\mathrm{3289}}}{\mathrm{11}} \\ $$$$\frac{{abc}}{\mathrm{4}{R}}=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{5}×\frac{\sqrt{\mathrm{120}}}{\mathrm{11}}\Rightarrow{R}=\frac{\sqrt{\mathrm{3289}}}{\:\mathrm{2}\sqrt{\mathrm{120}}} \\ $$
Commented by Tawa11 last updated on 18/Jun/24
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$