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Question-103928




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=((31)/(30))km  distance travel in y direction=(2/3)km  Net distance travelled=(√(((2/3))^2 +(((31)/(30)))^2 ))=1.51222222 km  Net angle θ=tan^(−1) ((20)/(31))=32.8285°
$${Total} \\ $$$${distance}\:{travel}\:{in}\:{x}\:{direction}=\frac{\mathrm{31}}{\mathrm{30}}{km} \\ $$$${distance}\:{travel}\:{in}\:{y}\:{direction}=\frac{\mathrm{2}}{\mathrm{3}}{km} \\ $$$${Net}\:{distance}\:{travelled}=\sqrt{\left(\frac{\mathrm{2}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{31}}{\mathrm{30}}\right)^{\mathrm{2}} }=\mathrm{1}.\mathrm{51222222}\:{km} \\ $$$${Net}\:{angle}\:\theta={tan}^{−\mathrm{1}} \frac{\mathrm{20}}{\mathrm{31}}=\mathrm{32}.\mathrm{8285}° \\ $$$$ \\ $$
Commented by I want to learn more last updated on 18/Jul/20
Thanks sir
$$\mathrm{Thanks}\:\mathrm{sir} \\ $$

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