(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.