Answer
$18.7\space N, 76^{\circ}$
Work Step by Step
Here we use the component method to find the $F_{3}$
Given that, the vector sum of the three forces is zero, we can write.
$\vec F_{1}+\vec F_{2}+\vec F_{3}=0$
$\vec F_{3}=-(\vec F_{1}+\vec F_{2})$
We can find the x component of F as follows.
$F_{3x}=-(F_{1x}+F_{2x})$ ; Let's plug known values into this equation.
$F_{3x}=-(-21\space N\times sin30^{\circ}+15\space N)=-4.5\space N$
We can find the y component of F as follows.
$F_{3y}=-(F_{1y}+F_{2y})$ ; Let's plug known values into this equation.
$F_{3y}=-(21\space N\times cos30^{\circ})=-18.2\space N$
By using the Pythagorean theorem, we can get.
$F_{3}=\sqrt {F_{3x}^{2}+F_{3y}^{2}}=\sqrt {(-4.5\space N)^{2}+(-18.2\space N)^{2}}=18.7\space N$
By using trigonometry, we can get.
$tan\theta=\frac{F_{3y}}{F_{3x}}=\frac{-18.2\space N}{-4.5\space N}=>\theta=tan^{-1}(4.04)=76^{\circ} $
