Answer
(a) $\langle -7.57, 10.61 \rangle$.
(b) $\langle 7.57, -10.61 \rangle$
Work Step by Step
(a) Given $\vec {F_1} =10cos60^{\circ} i+10sin60^{\circ} j=5i+5\sqrt 3 j$,
$\vec {F_2} =-8cos30^{\circ} i+8sin30^{\circ} j=-4\sqrt 3 i + 4 j$, and
$\vec {F_3} =-6cos20^{\circ} i-6sin20^{\circ} j\approx-5.64i-2.05j$,
the resultant force acting at P is the sum of the forces $\vec {F} =\vec {F_1} + \vec {F_2} + \vec {F_3} \approx-7.57i+10.61j$ or $\langle -7.57, 10.61 \rangle$.
(b) For the forces to be in equilibrium, the additional force required should be $\vec {F_a} \approx7.57i-10.61j$ or $\langle 7.57, -10.61 \rangle$ which satisfies the equation $\vec {F} + \vec {F_a} =0$
