To solve for θ and T, we resolve the forces vertically and horizontally.

Vertically:

\(T \sin \theta + 120 \sin 45^{\circ} = 15g\)

Horizontally:

\(T \cos \theta = 120 \cos 45^{\circ}\)

Using the vertical equation:

\(T \sin \theta = 15g - 120 \sin 45^{\circ}\)

Using the horizontal equation:

\(T = \frac{120 \cos 45^{\circ}}{\cos \theta}\)

Substitute for \(T\) in the vertical equation:

\(\frac{120 \cos 45^{\circ}}{\cos \theta} \sin \theta = 15g - 120 \sin 45^{\circ}\)

Using trigonometric identities and solving for \(\theta\):

\(\tan \theta = \frac{15g - 120 \sin 45^{\circ}}{120 \cos 45^{\circ}}\)

Calculate \(\theta\):

\(\theta = 37.5^{\circ}\)

Substitute \(\theta\) back to find \(T\):

\(T = \sqrt{(15g \sin \theta)^2 + (120 \cos 45^{\circ})^2}\)

Calculate \(T\):

\(T = 107 \text{ N}\)