Pushing a box

Kyle is pulling a box east with a force of 300 newtons at a constant angle of 42° to the horizontal.
Jerome is pushing the box from behind with a force of 350 newtons due east.
Determine the magnitude and direction of the resultant force on the box.

Let m = magnitude
Let D = direction

1. trigonometric:
$\begin{array}{|rcll|} \hline m^2 &=& 300^2+350^2-2\cdot 300 \cdot 350 \cdot \cos(180^{\circ} - 42^{\circ}) \\ m^2 &=& 212500+210000 \cdot \cos(42^{\circ}) \\ m^2 &=& 212500+210000 \cdot \cos(42^{\circ}) \\ m^2 &=& 212500+210000 \cdot 0.74314482548 \\ m^2 &=& 368560.413350 \\ \mathbf{m} & \mathbf{=} & \mathbf{607.091766828\ N } \\\\ \frac{\sin(D)}{300} &=& \frac{\sin(180^{\circ} - 42^{\circ})}{m} \\ \frac{\sin(D)}{300} &=& \frac{\sin(42^{\circ})}{m} \\ \sin(D) &=& \frac{300}{m} \cdot \sin(42^{\circ}) \\ \sin(D) &=& \frac{300}{m} \cdot \sin(42^{\circ}) \\ \sin(D) &=& \frac{300}{607.091766828} \cdot \sin(42^{\circ}) \\ \sin(D) &=& 0.49415922994 \cdot 0.66913060636 \\ \sin(D) &=& 0.33065706517 \\ \mathbf{D} & \mathbf{=} & \mathbf{19.3086615149^{\circ}} \\ \hline \end{array}$

2. vectorial:

$\begin{array}{|rcll|} \hline \vec{K} &=& \dbinom{350}{0} \\ \vec{J} &=& \dbinom{ 300\cdot \cos(42^{\circ}) } { 300\cdot \sin(42^{\circ}) } \\\\ \vec{K}+\vec{J} &=& \dbinom{350}{0} + \dbinom{ 300\cdot \cos(42^{\circ}) } { 300\cdot \sin(42^{\circ}) } \\ \vec{K}+\vec{J} &=& \dbinom{350+300\cdot \cos(42^{\circ})}{300\cdot \sin(42^{\circ})} \\ \vec{K}+\vec{J} &=& \dbinom{572.943447643}{200.739181908} \\\\ m &=& \sqrt{ 572.943447643^2+200.739181908^2} \\ \mathbf{m} & \mathbf{=} & \mathbf{607.091766828\ N } \\\\ \tan(D) &=& \frac{200.739181908}{572.943447643} \\ \tan(D) &=& 0.35036473972 \\ \mathbf{D} & \mathbf{=} & \mathbf{19.3086615149^{\circ}} \\ \hline \end{array}$