#### Answer

(a) The magnitude of the vector is 7.2 and the direction is an angle of $56.3^{\circ}$ below the positive x-axis.
(b) The magnitude of the vector is 94.3 m and the direction is an angle of $58^{\circ}$ above the positive x-axis.
(c) The magnitude of the vector is 44.7 m/s and the direction is an angle of $63.4^{\circ}$ above the negative x-axis.
(d) The magnitude of the vector is $6.3~m/s^2$ and the direction is an angle of $71.6^{\circ}$ below the positive x-axis.

#### Work Step by Step

(a) We can find the magnitude of the vector.
$A = \sqrt{A_x^2+A_y^2}$
$A = \sqrt{(4)^2+(-6)^2}$
$A = 7.2$
We can find the angle below the positive x-axis.
$tan(\theta) = \frac{6}{4}$
$\theta = tan^{-1}(\frac{6}{4}) = 56.3^{\circ}$
The magnitude of the vector is 7.2 and the direction is an angle of $56.3^{\circ}$ below the positive x-axis.
(b) We can find the magnitude of the vector.
$r = \sqrt{r_x^2+r_y^2}$
$r = \sqrt{(50~m)^2+(80~m)^2}$
$r = 94.3~m$
We can find the angle above the positive x-axis.
$tan(\theta) = \frac{80}{50}$
$\theta = tan^{-1}(\frac{80}{50}) = 58^{\circ}$
The magnitude of the vector is 94.3 m and the direction is an angle of $58^{\circ}$ above the positive x-axis.
(c) We can find the magnitude of the vector.
$v = \sqrt{v_x^2+v_y^2}$
$v = \sqrt{(-20~m/s)^2+(40~m/s)^2}$
$v = 44.7~m/s$
We can find the angle above the negative x-axis.
$tan(\theta) = \frac{40}{20}$
$\theta = tan^{-1}(\frac{40}{20}) = 63.4^{\circ}$
The magnitude of the vector is 44.7 m/s and the direction is an angle of $63.4^{\circ}$ above the negative x-axis.
(d) We can find the magnitude of the vector.
$a = \sqrt{a_x^2+a_y^2}$
$a = \sqrt{(2.0~m/s^2)^2+(-6.0~m/s^2)^2}$
$a = 6.3~m/s^2$
We can find the angle below the positive x-axis.
$tan(\theta) = \frac{6.0}{2.0}$
$\theta = tan^{-1}(\frac{6.0}{2.0}) = 71.6^{\circ}$
The magnitude of the vector is $6.3~m/s^2$ and the direction is an angle of $71.6^{\circ}$ below the positive x-axis.