Resolve each force into its x and y components.

• Force 1: $F_1 = 6\,N$ (acting vertically upwards, along the positive y-axis)

  • $F_{1x} = 6 \cos(90^\circ) = 0\,N$
  • $F_{1y} = 6 \sin(90^\circ) = 6\,N$

• Force 2: $F_2 = 10\,N$ (acting at $30^\circ$ above the positive x-axis)

  • $F_{2x} = 10 \cos(30^\circ) = 10 \left(\frac{\sqrt{3}}{2}\right) = 5\sqrt{3}\,N$
  • $F_{2y} = 10 \sin(30^\circ) = 10 \left(\frac{1}{2}\right) = 5\,N$

• Force 3: $F_3 = 3\,N$ (acting horizontally, along the positive x-axis)

  • $F_{3x} = 3 \cos(0^\circ) = 3\,N$
  • $F_{3y} = 3 \sin(0^\circ) = 0\,N$

• Force 4: $F_4 = 3\,N$ (acting at $60^\circ$ below the negative x-axis)

  • This force is in the third quadrant. Both x and y components will be negative.
  • $F_{4x} = -3 \cos(60^\circ) = -3 \left(\frac{1}{2}\right) = -1.5\,N$
  • $F_{4y} = -3 \sin(60^\circ) = -3 \left(\frac{\sqrt{3}}{2}\right) = -1.5\sqrt{3}\,N$

Numerically: $R_x \approx 5(1.732) + 1.5 = 8.66 + 1.5 = 10.16\,N$

Numerically: $R_y \approx 11 - 1.5(1.732) = 11 - 2.598 = 8.402\,N$