Question Number 131860 by aurpeyz last updated on 09/Feb/21
$${what}\:{is}\:{the}\:{magnitude}\:{and}\:\: \\ $$$${direction}\left({in}\:{degree}\right)\:{of}\:{this}\:{vector}? \\ $$$${F}=−\mathrm{3}×\mathrm{10}^{−\mathrm{6}} {i}−\mathrm{13}.\mathrm{35}×\mathrm{10}^{−\mathrm{6}} {j} \\ $$$$\left({a}\right)\:\mathrm{282}.\mathrm{5}^{\mathrm{0}} \:\left({b}\right)\mathrm{78}.\mathrm{5}^{\mathrm{0}} \:\left({c}\right)\:\mathrm{82}.\mathrm{5}^{\mathrm{0}} \:\left({d}\right)\mathrm{78}.\mathrm{5}^{\mathrm{0}} \\ $$$$\left({e}\right)\mathrm{282}.\mathrm{5}^{\mathrm{0}} \\ $$
Commented by mr W last updated on 09/Feb/21
$$\left({a}\right)\:{and}\:\left({e}\right)\:{are}\:{the}\:{same}. \\ $$$$\left({b}\right)\:{and}\:\left({d}\right)\:{are}\:{the}\:{same}. \\ $$$${but}\:{all}\:{wrong}! \\ $$$$\mathrm{257}.\mathrm{33}° \\ $$
Commented by Dwaipayan Shikari last updated on 09/Feb/21
$${tan}^{−\mathrm{1}} \frac{−\mathrm{13}.\mathrm{35}}{−\mathrm{3}}=\frac{\mathrm{180}}{\pi}{tan}^{−\mathrm{1}} \left(\mathrm{4}.\mathrm{43}\right)+\mathrm{180}°=\mathrm{257}.\mathrm{33}° \\ $$