this one is pretty basic.

Let $\vec{G}=(10,0)$ then $F=(8\cos(60\deg),8\sin(60\deg))=(4,4\sqrt{3})$

Now add them

$F+G=(14,4\sqrt{3})$

the length is given by $|F+G|=\sqrt{14^2+(4\sqrt{3})^2}=\sqrt{244}=2\sqrt {61}$

$\theta=\arctan\left(\dfrac{4\sqrt{3}}{14}\right)= 26.3\deg$