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If \(\vec A+\vec B=\vec C\) prove that \(C=(A^2+B^2+2AB\,cos\,\theta)^\frac{1}{2}\), where θ is the angle between A and B.

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\(\vec C=\vec A+\vec B\)

∴ \(\vec {C^2}=(\vec A+\vec B).(\vec A+\vec B)\)

or \(C^2=\vec A.\vec A+\vec B.\vec B+\vec A.\vec B+\vec B.\vec A\)

=\(A^2+B^2+2\vec A.\vec B\)

or C=\((A^2+B^2+2AB\,cos\,\theta)^\frac{1}{2}\)

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