We know 

$\sum F_x=50cos60^{\circ}+70cos45^{\circ}+25cos60^{\circ}\\ \sum F_x=86.99\approx 87N\\ \\ \\ \sum F_y=50sin60^{\circ}+70sin45^{\circ}-25sin60^{\circ}-20\\ \sum F_y=51.15\ N$

The resultant of the forces acting as given in above figure.

 $R=\sqrt{(\sum F_x)^2+(\sum F_y)^2}\\ R=\sqrt{(87)^2+(51.15)^2}\\ R=100.92\ N$

The angle which the resultant makes with the positive x-axis. 

$tan \theta =\dfrac{\sum F_y}{\sum F_x}\\ tan\theta =\dfrac{51.15}{87}\\ \theta =tan^{-1}\Bigg(\dfrac{51.15}{87}\Bigg)\\ \theta =30.45^{\circ}$