Answer
(a) $-22^{\circ}, 5.4m$
(b) $110^{\circ}, 5.4m$
(c) $45^{\circ}, 4.2m$
Work Step by Step
(a) We can find the direction as
$\theta=tan^{-1}(\frac{A_y}{A_x})$
$\theta=tan^{-1}(\frac{-2.0}{5.0})=-22^{\circ}$
and the magnitude can be determined as
$A=\sqrt{(x)^2+(y)^2}$
We plug in the known values to obtain:
$A=\sqrt{(5.0)^2+(-2.0)^2}=5.4m$
(b) We can find the direction as
$\theta=tan^{-1}(\frac{B_y}{B_x})$
$\theta=tan^{-1}(\frac{5.0}{-2.0})=110^{\circ}$
and the magnitude can be determined as
$B=\sqrt{(x)^2+(y)^2}$
We plug in the known values to obtain:
$B=\sqrt{(-2.0)^2+(5.0)^2}=5.4m$
(c) First of all, we find the x and y components of $\vec A+\vec B$
$\vec A+\vec B=(5.0-2.0m)\hat x+(-2.0+5.0m)\hat y=(3.0m)\hat x+(3.0m)\hat y$
Now the direction of the required vector is
$\theta=tan^{-1}(\frac{3.0}{3.0})=45^{\circ}$
and the magnitude is given as
$A+B=\sqrt{(3.0)^2+(3.0)^2}=4.2m$
