Question Number 198025 by mr W last updated on 08/Oct/23
Commented by mr W last updated on 08/Oct/23
$${prove}\:{PQ}={RS} \\ $$
Answered by mahdipoor last updated on 08/Oct/23
$${PQ}^{\mathrm{2}} ={MP}^{\mathrm{2}} +{MQ}^{\mathrm{2}} −\mathrm{2}.{MP}.{MQ}.{cosM} \\ $$$$=\mathrm{2}{R}_{{M}} ^{\mathrm{2}} \left(\mathrm{1}−{cosM}\right)=\mathrm{2}{R}_{{M}} ^{\mathrm{2}} \left(\mathrm{1}−\left(\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{{M}}{\mathrm{2}}\right)\right)= \\ $$$$\left(\mathrm{2}{R}_{{M}} {sin}\frac{{M}}{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{2}{R}_{{M}} ×\frac{{R}_{{N}} }{{MN}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\Rightarrow{PQ}=\frac{\mathrm{2}{R}_{{M}} {R}_{{N}} }{{MN}} \\ $$$${like}\:{that}\::\:{RS}=\mathrm{2}{R}_{{N}} {sin}\frac{{N}}{\mathrm{2}}=\frac{\mathrm{2}{R}_{{N}} {R}_{{M}} }{{MN}} \\ $$$$\Rightarrow\Rightarrow{RS}={PQ} \\ $$
Commented by mr W last updated on 09/Oct/23
$${thanks}! \\ $$