Decompose the forces acting on the rocket in horizontal and vertical components.
• net horizontal force:
∑ F = (12.3 N) cos(65°) - 2.5 N ≈ 2.7 N
(notice we're taking "to the right" to be the positive direction)
• net vertical force:
∑ F = (12.3 N) sin(65°) - 1.2 N ≈ 9.9 N
The resultant force then has magnitude
√((2.7 N)² + (9.9 N)²) ≈ 10.3 N
Since the horizontal and vertical components of the resultant are both positive, it points at an angle between 0° and 90° from the positive horizontal, so that
tan(θ) ≈ (9.9 N) / (2.7 N) ≈ 3.69
===> θ ≈ arctan(3.69) ≈ 75°
