91Ó°ÊÓ

It is given that $\stackrel{→}{A}=\left(3.0m,{20}^{∘}southofeast\right),\stackrel{→}{B}=\left(2.0mnorth\right)and\stackrel{→}{C}=\left(5.0m,{70}^{∘}southofwest\right).$We need to draw and label$\stackrel{→}{A},\stackrel{→}{B}and\stackrel{→}{C}$with the tails at the origin by using a coordinate system with the$x$-axis at the east.

The $\stackrel{→}{A},\stackrel{→}{B}and\stackrel{→}{C}$in the coordinate system is

It is given that$\stackrel{→}{A}=\left(3.0m,{20}^{∘}southofeast\right),\stackrel{→}{B}=\left(2.0mnorth\right)and\stackrel{→}{C}=\left(5.0m,{70}^{∘}southofwest\right).$We need to write$\stackrel{→}{A},\stackrel{→}{B}and\stackrel{→}{C}$in component form using unit vector.

$\stackrel{→}{A}=\left(3.0m,{20}^{∘}southofeast\right).$

$A_x=3m\cos \left(20\right).$

$=2.82m.$

$A_y=-3m\sin \left(20\right).$

$=-1.03m.$

$\stackrel{→}{A}=\left(2.82\hat{i}-1.03\hat{j}\right)m.$

$\stackrel{→}{B}=\left(2.0m,north\right).$

$B_x=2m\cos \left(90\right).$

$=0m.$

$B_y=2m\sin \left(90\right).$

$=1m.$

$B=\left(0\hat{i}+1\hat{j}\right)m.$

$\stackrel{→}{C}=\left(5.0m,{70}^{∘}southofwest\right).$

$C_x=-5m\cos \left(70\right).$

$=-1.71m.$

$C_y=-5m\sin \left(70\right).$

$=-4.70m.$

$\stackrel{→}{C}=\left(-1.71\hat{i}+4.70\hat{j}\right)m.$

It is given that,$\stackrel{→}{A}=\left(3.0m,{20}^{∘}southofeast\right),\stackrel{→}{B}=\left(2.0mnorth\right)and\stackrel{→}{C}=\left(5.0m,{70}^{∘}southofwest\right).$We need to determine the magnitude and direction ofrole="math" localid="1649269790122" $\stackrel{→}{D}=\stackrel{→}{A}+\stackrel{→}{B}+\stackrel{→}{C}.$

$\stackrel{→}{D}=\stackrel{→}{A}+\stackrel{→}{B}+\stackrel{→}{C}.$

$\stackrel{→}{D}=\left(A_x+B_x+C_x\right)\hat{i}+\left(A_y+B_y+C_y\right)\hat{j}.$

$=\left(2.82+0-1.71\right)m\hat{i}+\left(-1.03+2-4.70\right)m\hat{j}.$

$=\left(1.11\hat{i}+-3.73\hat{j}\right)m.$

$\left|\stackrel{→}{D}\right|=\sqrt{D_x^2+D_y^2}=\sqrt{1.{11}^2+{\left(-3.73\right)}^2}.$

$=3.89m.$

$\tan \mathrm{θ}=\frac{D_y}{D_x}=\frac{-3.73}{1.11}.$

$\mathrm{θ}={\tan }^{-1}\left(\frac{-3.73}{1.11}\right).$

$=-73.4^{∘}.$