#### Answer

$F_x = 451~N$
$F_y = 315~N$

#### Work Step by Step

Paul is pulling with a force $F$ that is directed at an angle of $\theta = 35^{\circ}$ above the ground.
We can find the horizontal component of the force;
$F_x = F~cos(\theta)$
$F_x = (550~N)~cos(35^{\circ})$
$F_x = 451~N$
We can find the horizontal component of the force;
$F_y = F~sin(\theta)$
$F_y = (550~N)~sin(35^{\circ})$
$F_y = 315~N$