Torque= R×F= (2.4 i^ + 3.9 j^)×(-5.3i^ + 3.2j^)= 21.438k^ Nm
a) magnitude= 21.438 Nm
b) we know that |R|•|F| sinø= 21.438
=>√(2.4^2 + 3.9^2) × √(5.3^2 + 3.2^2)×sinø=21.438
Hence sinø= 0.7561
Hence ø=49.1°
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