First, resolve the forces into their components along the i and j directions.

For the 31 N force, it acts entirely in the i direction: \(X_1 = 31\).

For the 26 N force, resolve using \(\tan \alpha = \frac{5}{12}\):

\(\cos \alpha = \frac{12}{13}\) and \(\sin \alpha = \frac{5}{13}\).

Thus, \(X_2 = 26 \cos \alpha = 26 \times \frac{12}{13} = 24\).

\(Y_2 = 26 \sin \alpha = 26 \times \frac{5}{13} = 10\).

For the 58 N force, it acts entirely in the j direction: \(Y_1 = 58\).

Sum the components:

\(X = 31 + 24 = 55\).

\(Y = 58 - 10 = 48\).

Calculate the magnitude of the resultant force:

\(R = \sqrt{X^2 + Y^2} = \sqrt{55^2 + 48^2} = 73 \text{ N}\).

Calculate the direction of the resultant force:

\(\tan \theta = \frac{Y}{X} = \frac{48}{55}\).

\(\theta = \arctan \left( \frac{48}{55} \right) \approx 41.1^\circ\).

Thus, the resultant is 73 N and the direction is 41.1° to the i direction.

(i) To find the component of the resultant in the direction of AB:

The forces are resolved along AB using the cosine of the angles given:

\(2 \times 12 \cos 40^\circ - 15 \cos 50^\circ = 8.74 \text{ N}\)

(ii) To find the component perpendicular to AB:

The forces are resolved perpendicular to AB using the sine of the angles:

\(12 \sin 40^\circ + 12 \sin 40^\circ - 15 \sin 50^\circ = 11.5 \text{ N}\)

(ii) To find the magnitude and direction of the resultant:

The magnitude of the resultant force is given by:

\(R = \sqrt{(8.74)^2 + (11.5)^2} = 14.4 \text{ N}\)

The direction \(\theta\) is given by:

\(\tan \theta = \frac{11.5}{8.74}\)

Thus, \(\theta = 52.7^\circ\) anticlockwise from the i direction.

To solve for \(F\) and \(\theta\), resolve the forces into horizontal and vertical components:

Horizontal: \(F \cos \theta = 12 \cos 30^\circ\)

Vertical: \(F \sin \theta = 10 - 12 \sin 30^\circ\)

Calculate \(F\) and \(\theta\):

\(F \cos \theta = 10.932\)

\(F \sin \theta = 4\)

Using \(F^2 = X^2 + Y^2\):

\(F = \sqrt{10.932^2 + 4^2} = 11.1 \text{ N}\)

Using \(\tan \theta = \frac{Y}{X}\):

\(\theta = \arctan \left( \frac{4}{10.932} \right) = 21.1^\circ\)

For part (ii), when the 12 N force is removed, the resultant force is the vector sum of the remaining forces:

Magnitude is 12 N, direction is 30° clockwise from the positive x-axis.

(i) To find \(F\), use the Pythagorean theorem for the components:

\(F^2 = 27.5^2 + (-24)^2\)

\(F = \sqrt{27.5^2 + 24^2} = 36.5\)

To find \(\alpha\), use the tangent function:

\(\tan \alpha = \frac{-24}{27.5}\)

\(\alpha = \arctan\left(\frac{-24}{27.5}\right) = 41.1^\circ\)

(ii) The second force acts at 90° anticlockwise, so the angle between the forces is \(\alpha + 90^\circ\). The magnitude of the resultant force \(R\) is given by:

\(R = \sqrt{F^2 + 87.6^2 + 2 \cdot F \cdot 87.6 \cdot \cos(90^\circ + \alpha)}\)

\(R = \sqrt{36.5^2 + 87.6^2 + 2 \cdot 36.5 \cdot 87.6 \cdot \cos(90^\circ + 41.1^\circ)} = 94.9\)

To find \(\theta\), use the tangent function:

\(\tan \theta = \frac{87.6 \cdot \sin(48.9^\circ) - 24}{27.5 + 87.6 \cdot \cos(48.9^\circ)}\)

\(\theta = \arctan\left(\frac{87.6 \cdot \sin(48.9^\circ) - 24}{27.5 + 87.6 \cdot \cos(48.9^\circ)}\right) = 26.3^\circ\)

To find \(\alpha\), we use the vertical component equation:

\(7.3 \sin \alpha = 5.5\)

Solving for \(\alpha\):

\(\sin \alpha = \frac{5.5}{7.3}\)

\(\alpha = \sin^{-1}\left(\frac{5.5}{7.3}\right) \approx 48.9^\circ\)

To find the magnitude of the resultant, we use the horizontal component equation:

\(R = 6.8 - 7.3 \cos 48.9^\circ\)

Calculating \(R\):

\(R \approx 6.8 - 7.3 \times 0.662\)

\(R \approx 6.8 - 4.835\)

\(R \approx 2 \text{ N}\)