(i) Since the particle is in equilibrium, the sum of the forces in both the horizontal and vertical directions must be zero.

Resolve the forces horizontally and vertically:

Horizontal: \(7 = F \cos \theta\)

Vertical: \(4 = F \sin \theta\)

Using \(F^2 = 7^2 + 4^2\), we find:

\(F = \sqrt{49 + 16} = \sqrt{65} \approx 8.06\)

Using \(\tan \theta = \frac{4}{7}\), we find:

\(\theta = \arctan\left(\frac{4}{7}\right) \approx 29.7^\circ\)

(ii) When the 7 N force is removed, the resultant force is the vector sum of the remaining forces:

The magnitude is 7 N, and the direction is opposite to the original 7 N force.

(i) To find \(\theta\), use the component of the resultant in the direction OA:

\(8 + 8 \cos \theta = 9\)

Solving for \(\cos \theta\):

\(8 \cos \theta = 1\)

\(\cos \theta = \frac{1}{8}\)

\(\theta = \cos^{-1} \left( \frac{1}{8} \right) \approx 82.8^\circ\)

(ii) To find the magnitude of the resultant, use the cosine rule in the triangle formed by the forces and the resultant:

\(R^2 = 8^2 + 8^2 + 2 \times 8 \times 8 \cos \theta\)

\(R^2 = 64 + 64 + 128 \cos \theta\)

Using \(\cos \theta = \frac{1}{8}\):

\(R^2 = 128 + 128 \times \frac{1}{8}\)

\(R^2 = 128 + 16\)

\(R^2 = 144\)

\(R = \sqrt{144} = 12 \text{ N}\)

(i) To find P, use the formula for the magnitude of a vector:

\(P = \sqrt{(-2.8)^2 + 9.6^2} = \sqrt{7.84 + 92.16} = \sqrt{100} = 10 \text{ N}\).

To find R, use the Pythagorean theorem for the resultant of perpendicular forces:

\(R = \sqrt{10^2 + 25^2} = \sqrt{100 + 625} = \sqrt{725} \approx 26.9 \text{ N}\).

(ii) To find α, use the tangent function:

\(\tan \alpha = \frac{9.6}{2.8}\), so \(\alpha = \arctan\left(\frac{9.6}{2.8}\right) \approx 73.7^\circ\).

For the components of the 25N force:

(a) x-direction: \(25 \cos(90^\circ - \alpha) = 25 \sin(\alpha) \approx 24 \text{ N}\).

(b) y-direction: \(25 \sin(90^\circ - \alpha) = 25 \cos(\alpha) \approx 7 \text{ N}\).

(iii) To find θ, use the tangent function for the resultant vector:

\(\tan \theta = \frac{7 + 9.6}{24 - 2.8}\), so \(\theta = \arctan\left(\frac{16.6}{21.2}\right) \approx 38.1^\circ\).

To find the resultant of the three forces, we resolve each force into its horizontal ( X ) and vertical ( Y ) components.

For the 7 N force (acting horizontally):

\(X_1 = 7, \quad Y_1 = 0\)

For the 5 N force (at 50° to the horizontal):

\(X_2 = 5 \cos 50°, \quad Y_2 = 5 \sin 50°\)

For the 6 N force (at 30° below the horizontal):

\(X_3 = -6 \cos 30°, \quad Y_3 = -6 \sin 30°\)

Summing the components:

\(X = X_1 + X_2 + X_3 = 7 + 5 \cos 50° - 6 \cos 30°\)

\(Y = Y_1 + Y_2 + Y_3 = 5 \sin 50° - 6 \sin 30°\)

Calculating these values:

X \approx 5.02, \quad Y \approx 0.83

The magnitude of the resultant force is given by:

\(R = \sqrt{X^2 + Y^2} \approx \sqrt{5.02^2 + 0.83^2} \approx 5.09 \text{ N}\)

The direction of the resultant force relative to the 7 N force is given by:

\(\theta = \tan^{-1} \left(\frac{Y}{X}\right) \approx \tan^{-1} \left(\frac{0.83}{5.02}\right) \approx 9.4°\)

Thus, the magnitude of the resultant force is 5.09 N, and its direction is 9.4° anti-clockwise from the force of magnitude 7 N.

To find the resultant force \(R\) and the angle \(\alpha\), we resolve the forces into their horizontal and vertical components.

Horizontal component, \(X = 100 + 250 \cos 70^\circ\)

Vertical component, \(Y = 300 - 250 \sin 70^\circ\)

Calculate \(X\) and \(Y\):

\(X = 100 + 250 \times 0.3420 = 185.5\)

\(Y = 300 - 250 \times 0.9397 = 65.1\)

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

\(R^2 = X^2 + Y^2\)

\(R^2 = 185.5^2 + 65.1^2\)

\(R = \sqrt{185.5^2 + 65.1^2} = 197\)

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

\(\tan \alpha = \frac{Y}{X} = \frac{65.1}{185.5}\)

\(\alpha = \arctan \left( \frac{65.1}{185.5} \right) = 19.3^\circ\)