(a) v appears to have a fixed direction along the positive x-axis. If ||u|| = 150 N, ||v|| = 220 N, then when θ = 30°, you have
u = (150 N) (cos(30°) i + sin(30°) j ) ≈ (129.904 i + 75 j ) N
v = (220 N) (cos(0°) i + sin(0°) j ) = (220 i ) N
(i and j are the unit vectors in the positive x and y directions)
and their sum is
u + v ≈ (349.904 i + 75 j ) N
with magnitude
||u + v|| ≈ √((349.904)² + (75)²) N ≈ 357.851 N ≈ 357.9 N
and at angle φ made with the positive x-axis such that
tan(φ) ≈ (75 N) / (349.904 N) → φ ≈ 12.098° ≈ 12.1°
(b) Letting θ vary from 0° to 180° would make v a function of θ :
u = (150 N) (cos(θ) i + sin(θ) j ) = (150 cos(θ) i + 150 sin(θ) j ) N
Then
u + v = ((220 + 150 cos(θ)) i + (150 sin(θ)) j ) N
→ M = ||u + v|| = √((220 + 150 cos(θ))² + (150 sin(θ))²) N
M = √(48,400 + 66,000 cos(θ) + 22,500 cos²(θ) + 22,500 sin²(θ)) N
M = 10 √(709 + 660 cos(θ)) N
(c) As a function of θ, u + v makes an angle α with the positive x-axis such that
tan(α) = (150 sin(θ) / (220 + 150 cos(θ))
→ α = tan⁻¹((15 sin(θ) / (22 + 15 cos(θ)))
(d) Filling in the table is just a matter of evaluating M and α for each of the given angles θ. For example, when θ = 0°,
M = 10 √(709 + 660 cos(0°)) N = 370 N
α = tan⁻¹((15 sin(0°) / (22 + 15 cos(0°))) = 0°
When θ = 30°, you get the same result as in part (a).
When θ = 60°,
M = 10 √(709 + 660 cos(60°)) N ≈ 323.3 N
α = tan⁻¹((15 sin(60°) / (22 + 15 cos(60°))) = 23.8°
and so on.