# If vector A + vector B = vector C prove that C = (A^2 + B^2 + 2AB cos θ)^1/2, where θ is the angle between A and B.

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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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