Question Number 49006 by peter frank last updated on 01/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 01/Dec/18
$$\overset{\rightarrow} {{V}}={iV}_{{x}} +{jV}_{{y}} =\mathrm{12}{i}+\mathrm{8}{j} \\ $$$$\mid\overset{\rightarrow} {{V}}\mid=\sqrt{\left(\mathrm{12}\right)^{\mathrm{2}} +\left(\mathrm{8}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{144}+\mathrm{64}}\:=\sqrt{\mathrm{208}}\:=\sqrt{\mathrm{4}×\mathrm{4}×\mathrm{13}}\:=\mathrm{4}\sqrt{\mathrm{13}}\: \\ $$$${tan}\theta=\frac{\mathrm{8}}{\mathrm{12}}.=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\overset{\rightarrow} {{R}}={iR}_{{x}} +{jR}_{{y}} =−\mathrm{18}{i}+\mathrm{12}{j} \\ $$$$\mid\overset{\rightarrow} {{R}}\mid=\sqrt{\left(−\mathrm{18}\right)^{\mathrm{2}} +\left(\mathrm{12}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{324}+\mathrm{144}}\:=\sqrt{\mathrm{468}} \\ $$$$\:\sqrt{\mathrm{4}×\mathrm{117}}\:=\sqrt{\mathrm{4}×\mathrm{3}×\mathrm{39}}\:=\sqrt{\mathrm{4}×\mathrm{3}^{\mathrm{2}} ×\mathrm{13}}\:=\mathrm{6}\sqrt{\mathrm{13}}\: \\ $$$${tan}\alpha=\frac{\mathrm{12}}{−\mathrm{18}}=\frac{−\mathrm{2}}{\mathrm{3}} \\ $$$$\overset{\rightarrow} {{V}}+\overset{\rightarrow} {{R}}={i}\left(\mathrm{12}−\mathrm{18}\right)+{j}\left(\mathrm{8}+\mathrm{12}\right)=−\mathrm{6}{i}+\mathrm{20}{j} \\ $$$$\mid\overset{\rightarrow} {{V}}+\overset{\rightarrow} {{R}}\mid=\sqrt{\left(−\mathrm{6}\right)^{\mathrm{2}} +\left(\mathrm{20}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{436}}\:=\sqrt{\mathrm{4}×\mathrm{109}}\: \\ $$$${tan}\beta=\frac{\mathrm{20}}{−\mathrm{6}}=\frac{−\mathrm{10}}{\mathrm{3}} \\ $$$$ \\ $$
Commented by peter frank last updated on 01/Dec/18
$$\mathrm{thank}\:\mathrm{you}\: \\ $$