Answer
Magnitude = 4788 N, Direction = $67.2^{\circ}$ south of east
Work Step by Step
Here we use the component method to find the resultant force.
The x component of the resultant force $F$ is,
$F_{x}= 2240\space N\times cos34^{\circ}=2240\times 0.83=1857\space N$
The y component of the resultant force $F$ is,
$F_{y}=-2240\space N\times sin34^{\circ}+(-3160\space N)=-4413\space N$
Using the Pythagorean theorem, we can get
$F=\sqrt {(F_{x})^{2}+(F_{y})^{2}}=\sqrt {(1857\space N)^{2}+(-4413\space N)^{2}}=4788\space N$
Using trigonometry, we can get
$tan\theta=(\frac{4413\space N}{1857\space N})=\gt\theta=tan^{-1}(2.4)=67.2^{\circ}$
