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Question Number 49006 by peter frank last updated on 01/Dec/18

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Dec/18

$$\overrightarrow{V}={iV}_x+{jV}_y=12i+8j$$$$\mid \overrightarrow{V}\mid =\sqrt{{\left(12\right)}^2+{\left(8\right)}^2}=\sqrt{144+64}=\sqrt{208}=\sqrt{4\times 4\times 13}=4\sqrt{13}$$$$tan\theta =\frac{8}{12}.=\frac{2}{3}$$$$\overrightarrow{R}={iR}_x+{jR}_y=-18i+12j$$$$\mid \overrightarrow{R}\mid =\sqrt{{(-18)}^2+{\left(12\right)}^2}=\sqrt{324+144}=\sqrt{468}$$$$\sqrt{4\times 117}=\sqrt{4\times 3\times 39}=\sqrt{4\times 3^2\times 13}=6\sqrt{13}$$$$tan\alpha =\frac{12}{-18}=\frac{-2}{3}$$$$\overrightarrow{V}+\overrightarrow{R}=i(12-18)+j(8+12)=-6i+20j$$$$\mid \overrightarrow{V}+\overrightarrow{R}\mid =\sqrt{{(-6)}^2+{\left(20\right)}^2}=\sqrt{436}=\sqrt{4\times 109}$$$$tan\beta =\frac{20}{-6}=\frac{-10}{3}$$$$$$

Commented by peter frank last updated on 01/Dec/18

$$\mathrm{thank}\mathrm{you}$$

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