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b = 3
θ = 60°
ditanya : |a-b| = .... ?
dijawab : |a-b|² = a² + (-b)² + 2ab cos θ
|a-b|² = 5² + (-3)² + (2x5x(-3) x cos (60⁰))
= 25 + 9 - (30 x (1/2))
= 34 - 15
|a-b|² = 19
|a-b| = √19 satuan