Need Help Solving Question

direction angle = $\varphi$

magnitude = m

$\begin{array}{rcll} \vec{F}_1 &=& -100\cdot \binom{ \sin{(25^{\circ})} }{ \cos{(25^{\circ})} } \\ \vec{F}_1 &=& -200\cdot \binom{ \sin{(80^{\circ})} }{ \cos{(80^{\circ})} } \\ \vec{F} &=& \vec{F}_1+ \vec{F}_2 \\ \vec{F} &=& -100\cdot \dbinom{ \sin{(25^{\circ})} }{ \cos{(25^{\circ})} }-200\cdot \dbinom{ \sin{(80^{\circ})} }{ \cos{(80^{\circ})} }\\ \vec{F} &=& -\left[ \dbinom{ 100\cdot \sin{(25^{\circ})}+ 200\cdot \sin{(80^{\circ})} }{ 100\cdot \cos{(25^{\circ})} +200\cdot \cos{(80^{\circ})}} \right] \\ \vec{F} &=& -\left[ \dbinom{ 100\cdot 0.42261826174+ 200\cdot 0.98480775301 }{ 100\cdot 0.90630778704 +200\cdot 0.17364817767 } \right] \\ \vec{F} &=& -\left[ \dbinom{ 42.2618261741+ 196.961550602 }{ 90.6307787037 + 34.7296355334 } \right] \\ \vec{F} &=& -\left[ \dbinom{ 239.223376777 }{ 125.360414237 } \right] \\\\ \tan{ ( \varphi ) } &=& \frac{239.223376777}{125.360414237} \\ \varphi &=& 62.3440776108^{\circ}\\\\ m &=& \sqrt{239.223376777^2+ 125.360414237^2 }\\ m &=& 270.079724256\\\\ \vec{F} &=& -m \cdot \dbinom{ \sin{(\varphi)} } { \cos{(\varphi)} } \\ \vec{F} &=& -270.079724256 \cdot \dbinom{ \sin{(62.3440776108^{\circ})} }{ \cos{(62.3440776108^{\circ})} } \end{array}$

The magnitude is 270 pounds.

The direction is N$62.3^{\circ}$E