Answer
(a) $-26^{\circ}, 28m$
(b) $82^{\circ}, 15m$
(c) $6.3^{\circ},27m$
Work Step by Step
(a) We can find the direction as
$\theta=tan^{-1}(\frac{A_y}{A_x})$
$\theta=tan^{-1}(\frac{-12}{25m})=-26^{\circ}$
and the magnitude can be determined as
$A=\sqrt{(x)^2+(y)^2}$
We plug in the known values to obtain:
$A=\sqrt{(25)^2+(-12)^2}=28m$
(b) We can find the direction as
$\theta=tan^{-1}(\frac{B_y}{B_x})$
$\theta=tan^{-1}(\frac{15}{2.0})=82^{\circ}$
and the magnitude can be determined as
$B=\sqrt{(x)^2+(y)^2}$
We plug in the known values to obtain:
$B=\sqrt{(2.0)^2+(15)^2}=15m$
(c) First of all, we find the x and y components of $\vec A+\vec B$
$\vec A+\vec B=(25-2.0m)\hat x+(-12+15m)\hat y=(27m)\hat x+(3.0m)\hat y$
Now the direction of the required vector is
$\theta=tan^{-1}(\frac{3.0}{27})=6.3^{\circ}$
and the magnitude is given as
$A+B=\sqrt{(27)^2+(3.0)^2}=27m$
