Question Number 220231 by fantastic last updated on 09/May/25
Answered by mr W last updated on 09/May/25
$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}.\mathrm{5}}{\mathrm{1}+\mathrm{1}.\mathrm{5}+\sqrt{\left(\mathrm{1}+\mathrm{1}.\mathrm{5}\right)^{\mathrm{2}} −\mathrm{1}.\mathrm{5}^{\mathrm{2}} }}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$?=\mathrm{2}{R}=\frac{\mathrm{2}×\mathrm{1}.\mathrm{5}}{\mathrm{sin}\:\mathrm{2}\alpha}=\frac{\mathrm{3}×\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\right)}{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{3}}}=\mathrm{5}\:\checkmark \\ $$
Commented by fantastic last updated on 09/May/25
$${please}\:{in}\:{details} \\ $$
Commented by mr W last updated on 09/May/25
$${see}\:{below} \\ $$
Answered by mr W last updated on 09/May/25
Commented by mr W last updated on 09/May/25
$${a}=\frac{\mathrm{2}}{\mathrm{2}}=\mathrm{1},\:{b}=\frac{\mathrm{3}}{\mathrm{2}}=\mathrm{1}.\mathrm{5} \\ $$$${R}={a}+{b} \\ $$$$?=\mathrm{2}{R}=\mathrm{2}\left({a}+{b}\right)=\mathrm{2}+\mathrm{3}=\mathrm{5} \\ $$
Commented by fantastic last updated on 10/May/25
$${thanks} \\ $$
Answered by Spillover last updated on 10/May/25
Answered by mehdee7396 last updated on 10/May/25
$${AM}=\mathrm{2}{r}'{cos}\alpha\:\:\:\&\:\:\:\:{MB}=\mathrm{2}{rcos}\alpha \\ $$$$\Rightarrow{AB}=\mathrm{2}\left({r}+{r}'\right){cos}\alpha \\ $$$$\mathrm{2}{R}=\frac{{AB}}{{cos}\alpha}=\mathrm{2}\left({r}+{r}'\right)\:\:\checkmark\: \\ $$$$ \\ $$
Commented by mehdee7396 last updated on 10/May/25
