They were expected to resolve the forces vertically and horizontally to have; (${}_{5\sin {60}^o}^{5\cos {60}^o}$) + (${}_{P\sin {30}^o}^{P\cos {30}^o}$) + (${}_{-Q\sin {60}^o}^{Q\cos {60}^o}$) + (${}_{-3\sin {90}^o}^{3\cos {90}^o}$) + (${}_{5\sin 0^o}^{-5\cos 0^o}$) = (${}_0^0$)
which simplifies to;
(${}_{0.5P-0.866Q}^{0.866P+0.5Q}$) = (${}_{-1.33}^{2.5}$)
This will yield two equations to be solved, the equations are 0.866P + 0.5Q = 2.5 and 0.5P - 0.866Q = -1.33
When these equations are solved, they will give P = 1.5 and Q = 2.4019