Vector A has a magnitude of 3 m and is directed at an angle of 20^...

The angle between vectors $\vec{A}$ and $\vec{B}$ is \theta = 110^{}^{{}^{\circ}} - 20^{}^{{}^{\circ}} = 90^{}^{{}^{\circ}}.

Since the vectors are perpendicular, the magnitude of the resultant $\vec{R}$ is given by:

$R = \sqrt{A^2 + B^2} = \sqrt{3^2 + 4^2} = 5 \text{ m}$

The angle $\alpha$ that the resultant makes with vector $\vec{A}$ is:

\tan \alpha = \dfrac{B}{A} = \dfrac{4}{3} \Rightarrow \alpha = 53^{}^{{}^{\circ}}

The angle of the resultant with the X-axis is the sum of the angle of $\vec{A}$ with the X-axis and the angle $\alpha$:

\theta_R = 20^{}^{{}^{\circ}} + 53^{}^{{}^{\circ}} = 73^{}^{{}^{\circ}}

Thus, the resultant is $5 \text{ m}$ at an angle of 73^{}^{{}^{\circ}} with the X-axis.