Question Number 103928 by I want to learn more last updated on 18/Jul/20
Answered by Dwaipayan Shikari last updated on 18/Jul/20
$${Total} \\ $$$${distance}\:{travel}\:{in}\:{x}\:{direction}=\frac{\mathrm{31}}{\mathrm{30}}{km} \\ $$$${distance}\:{travel}\:{in}\:{y}\:{direction}=\frac{\mathrm{2}}{\mathrm{3}}{km} \\ $$$${Net}\:{distance}\:{travelled}=\sqrt{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{31}}{\mathrm{30}}\right)^{\mathrm{2}} }=\mathrm{1}.\mathrm{51222222}\:{km} \\ $$$${Net}\:{angle}\:\theta={tan}^{−\mathrm{1}} \frac{\mathrm{20}}{\mathrm{31}}=\mathrm{32}.\mathrm{8285}° \\ $$$$ \\ $$
Commented by I want to learn more last updated on 18/Jul/20
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$