Answer
$146^{\circ}, 7.8N$
Work Step by Step
We know that
$F_{1x}+F_{2x}+F_{3x}=0$
$\implies 6.5N+0+F_{3x}=0$
$\implies F_{3x}=-6.5N$
Similarly, $F_{1y}+F_{2y}+F_{3y}=0$
$\implies 0-4.4N+F_{3y}=0$
$F_{3y}=4.4N$
Now $F_3=\sqrt{F_{3x}^2+F_{3y}^2}$
We plug in the known values to obtain:
$F_3=\sqrt{(-6.5N)^2+(4.4N)^2}$
$F_3=7.8N$
We can find the direction of the force as
$\theta=tan^{-1}(\frac{F_{3y}}{F_{3x}})$
We plug in the known values to obtain:
$\theta=tan^{-1}(\frac{4.4N}{-6.5N})$
$\theta=146^{\circ}$
