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Verified answer
VEKTOR• komponen
a][
Fx = 5√3 N
Fy = - 5 N
besar
F = √(Fx² + Fy²)
F = √((5√3)² + (-5)²)
F = √100 = 10 N
arah
tan α = Fy / Fx
tan α = -5 / 5√3 (kuadran IV)
tan α = - ⅓√3
α = -30° = 330° ← jwb
b][
Fx = 4√3 N
Fy = 4 N
besar
F = √(Fx² + Fy²)
F = √((4√3)² + 4²)
F = √64 = 8 N
arah
tan α = Fy / Fx
tan α = 4 / 4√3 (kuadran I)
tan α = ⅓√3
α = 30° ← jwb