Answer
The angle between $F_1$ and $F_2$ is $~~56.3^{\circ}$
Work Step by Step
The slope of the $v_x$ versus $t$ curve is the horizontal component of acceleration $a_x$
We can find $a_x$:
$a_x = \frac{6.0~m/s}{2.0~s} = 3.0~m/s^2$
We can find $F_x$, the horizontal component of the net force:
$F_x = m~a_x$
$F_x = (4.0~kg)(3.0~m/s^2)$
$F_x = 12~N$
We can find the horizontal component of $F_2$:
$F_1+F_{2x} = 12~N$
$F_{2x} = 12~N - F_1$
$F_{2x} = 12~N - 7.0~N$
$F_{2x} = 5.0~N$
We can find the direction of $F_2$ as an angle $\theta$ relative to the x axis:
$F_2~cos~\theta = 5.0~N$
$cos~\theta = \frac{5.0~N}{F_2}$
$\theta = cos^{-1}~(\frac{5.0~N}{F_2})$
$\theta = cos^{-1}~(\frac{5.0~N}{9.0~N})$
$\theta = 56.3^{\circ}$
The angle between $F_1$ and $F_2$ is $~~56.3^{\circ}$
