Solution
Magnitude of vector A= 6 m
Magnitude of vector B=4 m
Angle between A and B vectors
θ=30°\theta=30^°θ=30°
So resultant vector
R=A2+B2+2ABcosθR=\sqrt{A^2+B^2+2AB\cos\theta}R=A2+B2+2ABcosθ
R=62+42+2×6×4cos30°R=9.67mR=\sqrt{6^2+4^2+2\times6\times4\cos 30^°}\\R=9.67mR=62+42+2×6×4cos30°R=9.67m
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