Question Number 212982 by Spillover last updated on 27/Oct/24
Answered by ajfour last updated on 28/Oct/24
$$\mathrm{2}{r}=\frac{\mathrm{85}}{\mathrm{3}}\:\: \\ $$
Answered by A5T last updated on 27/Oct/24
$$\frac{\mathrm{1}}{\mathrm{2}}×{a}×{bsin}\left(\mathrm{90}+\theta\right)=\frac{{abc}}{\mathrm{4}{R}}\:{where}\:{a}=\sqrt{\mathrm{8}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }=\mathrm{17} \\ $$$${c}=\sqrt{\mathrm{20}^{\mathrm{2}} +\mathrm{15}^{\mathrm{2}} }=\mathrm{25};{sin}\left(\mathrm{90}+\theta\right)=\frac{\mathrm{15}}{\mathrm{17}} \\ $$$$\Rightarrow{R}=\frac{{c}}{\mathrm{2}{cos}\theta}=\frac{\mathrm{25}}{\mathrm{2}×\frac{\mathrm{15}}{\mathrm{17}}}=\frac{\mathrm{5}×\mathrm{17}}{\mathrm{6}}=\frac{\mathrm{85}}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{2}{R}=\frac{\mathrm{85}}{\mathrm{3}} \\ $$
Commented by A5T last updated on 27/Oct/24
