Use the given information to draw a sketch of the situation. By imagining or sketching the parallelogram rule, it should be apparent that the resultant vector points up and to the left.
\begin{align*}
A_x \amp = \N{200} \cos \ang{45} = \N{141.4} \amp B_x \amp = - \N{300} \sin \ang{70}= \N{-281.9}\\
A_y \amp = \N{200} \sin \ang{45} = \N{141.4}\amp B_y \amp = \N{300} \cos \ang{70} = \N{102.6}\\
\\
R_x \amp = A_x + B_x \amp R_y \amp = A_y + B_y\\
\amp = \N{141.4} - \N{281.9}\amp \amp = \N{141.4} + \N{102.6}\\
\amp = \N{-140.5} \amp \amp = \N{244.0}\\
\\
R \amp = \sqrt{R_x^2 + R_y^2} = \N{281.6} \amp \theta \amp= \tan^{-1}\left( \frac{R_y}{R_x} \right) = \ang{-60.1}
\end{align*}
This answer indicates that the resultant points down and to the left, which is odd because the parallelogram rule shows that the resultant should point up and to the left.
This occurs because the calculator always returns angles in the first or fourth quadrant for \(\tan^{-1}\text{.}\) To get the actual direction of the resultant, add \(\ang{180}\) to the calculator result.
\begin{equation*}
\theta = \ang{-60.1} + \ang{180} = \ang{119.9}
\end{equation*}
The final answer for the magnitude and direction of the resultant is
\begin{equation*}
\vec{R} = \N{281.6} \angle \ang{119.9}
\end{equation*}
measured counter-clockwise from the \(x\) axis.