(i) Resolve the forces in the \(x\)-direction:

\(X = 25 \times 0.96 - 30 \times 0.8 = 0\)

Thus, the component in the \(x\)-direction is zero.

(ii) Resolve the forces in the \(y\)-direction:

\(Y = 25 \times 0.28 - 20 + 30 \times 0.6 = 5\)

The resultant has a magnitude of 5 N and acts in the positive \(y\) direction.

(iii) To reverse the direction of the resultant force, the replacement force must have a magnitude of 30 N and act in the negative \(y\) direction.

Resolve the forces into horizontal ( X ) and vertical ( Y ) components:

Horizontal component ( X ):

\(X = 20 \, \cos 60° - 14 - 16 \, \cos 50° = -14.2846 \, \text{N}\)

Vertical component ( Y ):

\(Y = 60 - 20 \, \sin 60° - 16 \, \sin 50° = 30.42278 \, \text{N}\)

Calculate the magnitude of the resultant force ( R ):

\(R = \sqrt{(-14.2846)^2 + (30.42278)^2} = 33.6 \, \text{N}\)

Calculate the direction of the resultant force:

\(\theta = \tan^{-1}\left(\frac{30.42278}{14.2846}\right) = 64.8°\)

The direction is 64.8° above the 14 N force.

(i) To find the y-components of the forces, use the Pythagorean theorem:

\(Y_1^2 = 68^2 - (-60)^2\)

\(Y_1 = \sqrt{68^2 - 60^2} = 32\)

\(Y_2 = 75\) (since the force is vertical)

\(Y_3^2 = 100^2 - 96^2\)

\(Y_3 = \sqrt{100^2 - 96^2} = 28\)

Thus, the components in the positive y-direction are -32, 75, and -28.

(ii) The resultant force components are:

\(X = -60 + 0 + 96 = 36\)

\(Y = -32 + 75 - 28 = 15\)

The magnitude of the resultant force is:

\(R = \sqrt{X^2 + Y^2} = \sqrt{36^2 + 15^2} = 39 \text{ N}\)

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

\(\theta = \arctan \left( \frac{Y}{X} \right) = \arctan \left( \frac{15}{36} \right) \approx 22.6^\circ\)

The direction is 22.6° anticlockwise from the positive x-axis.

Resolve the forces into their components along the x and y axes.

For the x-direction:

\(X = 12\cos 25^{\circ} - 8\cos 10^{\circ} = 2.9972\)

For the y-direction:

\(Y = 12\sin 25^{\circ} + 8\sin 10^{\circ} - 2 = 4.4606\)

The magnitude of the resultant force \(R\) is given by:

\(R = \sqrt{X^2 + Y^2} = \sqrt{(2.9972)^2 + (4.4606)^2} = 5.37\)

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

\(\tan \theta = \frac{X}{Y} = \frac{2.9972}{4.4606}\)

\(\theta = \arctan\left(\frac{2.9972}{4.4606}\right) = 33.9^{\circ}\)

(i) Using the cosine rule in the triangle formed by the forces:

\(X = 14 - 13 \cos \theta\)

\(Y = 13 \sin \theta\)

\(14^2 + 13^2 - 2 \times 13 \times 14 \cos \theta = 15^2\)

Solving for \(\theta\), we find \(\theta = 67.4^\circ\).

(ii) The component of the resultant in the direction of the 14 N force is:

\(X = 15 \cos \left( \arctan \left( \frac{Y}{X} \right) \right)\)

The component is 9 N.