WAEC Further Mathematics 2013 Paper

Concept:For a body to remain at rest, the vector sum of all forces acting on it must be zero.So the third force $F_3$ must be equal in magnitude and opposite in direction to $F_1 + F_2$.Explanation:Resolve $F_1 = (10\,\mathrm{N}, 030^\circ)$ into horizontal and vertical components.$F_{1x} = 10\cos 30^\circ = 5\sqrt{3}\,\mathrm{N}$.$F_{1y} = 10\sin 30^\circ = 5\,\mathrm{N}$.Resolve $F_2 = (10\,\mathrm{N}, 150^\circ)$ into components.$F_{2x} = 10\cos 150^\circ = -5\sqrt{3}\,\mathrm{N}$.$F_{2y} = 10\sin 150^\circ = 5\,\mathrm{N}$.Add the components of $F_1$ and $F_2$.Horizontal: $5\sqrt{3} + (-5\sqrt{3}) = 0$.Vertical: $5 + 5 = 10\,\mathrm{N}$.Thus, $F_1 + F_2 = (0, 10)\,\mathrm{N}$, which is $10\,\mathrm{N}$ at $090^\circ$.Since the body is at rest, $F_3$ must balance this resultant completely.Therefore, $F_3$ acts in the opposite direction with the same magnitude.Answer:$F_3 = (10\,\mathrm{N}, 270^\circ)$