To solve this problem, we resolve the forces into components along and perpendicular to the direction of PQ.

(i) (a) For resolving in the direction PQ:

The component is given by:

\(2 \times 10 \cos 30^\circ - 6 \cos 60^\circ\)

\(= 20 \times \frac{\sqrt{3}}{2} - 6 \times \frac{1}{2}\)

\(= 10\sqrt{3} - 3\)

\(\approx 14.3 \text{ N}\)

(i) (b) For resolving perpendicular to PQ:

The component is given by:

\(\pm 6 \cos 30^\circ - 6 \cos 60^\circ\)

\(= \pm 6 \times \frac{\sqrt{3}}{2} - 6 \times \frac{1}{2}\)

\(= \pm 3\sqrt{3} - 3\)

\(\approx \pm 5.20 \text{ N}\)

(ii) To find the magnitude of the resultant of the three forces:

Use the formula:

\(\sqrt{(\text{component in direction of } PQ)^2 + (\text{component perpendicular to } PQ)^2}\)

\(= \sqrt{(14.3)^2 + (5.20)^2}\)

\(\approx 15.2 \text{ N}\)

First, resolve the forces horizontally and vertically. Let the horizontal component be \(X\) and the vertical component be \(Y\).

For the horizontal component:

\(X = 30 - 34 \times \frac{8}{17} - 26 \times \frac{5}{13} = 4\)

For the vertical component:

\(Y = 34 \times \frac{15}{17} - 26 \times \frac{12}{13} = 6\)

Now, calculate the magnitude of the resultant force \(R\):

\(R = \sqrt{X^2 + Y^2} = \sqrt{4^2 + 6^2} = \sqrt{52} = 2\sqrt{13} = 7.21 \text{ N}\)

To find the direction \(\beta\) of the resultant force, use:

\(\beta = \arctan \left( \frac{Y}{X} \right) = \arctan \left( \frac{6}{4} \right) = 56.3^\circ\)

The direction is 56.3° above the 30 N force or anticlockwise from the 30 N force.

(a) Resolve forces horizontally:

\(20 \cos 30 = 25 \cos 60 + 10 \cos \alpha\)

\(17.32 = 12.5 + 10 \cos \alpha\)

\(\cos \alpha = 0.4821\)

\(\alpha = 61.2\)

Resolve forces vertically:

Resultant = \(20 \sin 30 + 10 \sin 61.2 - 25 \sin 60\)

= \(10 + 8.761 - 21.651\)

\(Magnitude of resultant force = 2.89 N\)

(b) Resolve forces horizontally and vertically:

\(X = 25 \cos 60 + 10 \cos 45 - 20 \cos 30\)

= \(12.5 + 7.07107 - 17.32051 = 2.25056\)

\(Y = 20 \sin 30 + 10 \sin 45 - 25 \sin 60\)

= \(10 + 7.07107 - 21.65064 = -4.57957\)

Resultant \(R = \sqrt{X^2 + Y^2}\)

= \(\sqrt{2.25056^2 + (-4.57957)^2} = 5.10 \text{ N}\)

Direction \(\alpha = \arctan \left( \frac{Y}{X} \right)\)

= \(\arctan \left( \frac{-4.57957}{2.25056} \right) = 63.8^\circ\) below positive x-axis

The forces acting at point A are 100 N to the left, and two 50 N forces at angles \(\alpha\) to the horizontal. The horizontal component of each 50 N force is \(50 \cos \alpha\).

The total horizontal component of the two 50 N forces is \(2 \times 50 \cos \alpha = 100 \cos \alpha\).

Given \(\cos \alpha = \frac{4}{5}\), the total horizontal component becomes \(100 \times \frac{4}{5} = 80\) N.

The resultant force is the difference between the 100 N force and the total horizontal component of the two 50 N forces:

\(100 - 80 = 20 \text{ N}\)

The direction of the resultant force is to the left.

(a) Resolve the forces vertically and equate to zero:

\(4 \sin 30^\circ + F \sin 60^\circ - 6 = 0\)

Substitute \(\sin 30^\circ = 0.5\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):

\(2 + \frac{F \sqrt{3}}{2} - 6 = 0\)

\(\frac{F \sqrt{3}}{2} = 4\)

\(F = \frac{8}{\sqrt{3}} \approx 4.62\)

(b) Resolve forces either vertically or horizontally:

Vertically: \(F \sin \alpha + 4 \sin 30^\circ - 6 = 0\)

Horizontally: \(F \cos \alpha + 3 - 4 \cos 30^\circ = 0\)

\(F \sin \alpha = 4\)

\(F \cos \alpha = 0.464\)

Using Pythagoras:

\(F^2 = 4^2 + 0.464^2\)

\(F = \sqrt{16 + 0.215296} \approx 4.03\)

Using trigonometry:

\(\alpha = \arctan \left( \frac{4}{0.464} \right) \approx 83.4\)