Answer
The magnitude and direction of the third leg is $2.81km$ ,$61.7^{\circ}$ North of East.
Work Step by Step
We take our coordinate system such that the (+) y-direction is along the North and (+) x-direction is along the East.
Given:$A=2.00km$ due East
$B=3.50km$ SouthEast
$C=?$(our unknow vector)
$R=5.80km$ due East(resultant vector)
We know, $R=A+B+C$
$=>C=R-A-B$
So,
$C_{x}=R_{x}-A_{x}-B_{x}=5.80cos(0^{\circ})-2.00cos(0^{\circ})-3.50cos(-45.0^{\circ})$=1.33km
and,$C_{y}=R_{y}-A_{y}-B_{y}=5.80sin(0^{\circ})-2.00sin(0^{\circ})-3.50sin(-45.0^{\circ})$=2.47km
Therefore,$C=(1.33km)i+(2.47km)j$ and lies in the 1st quadrant.
The magnitude of $C$ is
$|C|=\sqrt (1.33^2+2.47^2)$=2.81km
and the direction of $C$ is given by
$tan(\theta)=\frac{C_{y}}{C_{x}}=\frac{2.47}{1.33}$
$\theta=tan^{-1}(\frac{2.47}{1.33})=61.7^{\circ}$
Thus,the magnitude and direction of the third leg is $2.81km$ ,$61.7^{\circ}$ North of East.
From the vector-addition diagram approximately to scale,we get
$|C|\approx2.80km$
$\theta\approx60^{\circ}$North of East
Therefore ,these values are in qualitative agreement with our numerical solutions.
