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In any triangle ABC,Prove by vector method c^2=a^2+b^2-2abcosC​

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xero099

Answer:

Let be [tex]\vec A[/tex], [tex]\vec B[/tex] and [tex]\vec C[/tex] the vector of a triangle so that [tex]\vec C = \vec A + \vec B[/tex]. By definition of Dot Product:

[tex]\vec C \,\bullet\,\vec C = (\vec A + \vec B) \,\bullet \vec C[/tex]

[tex]\vec C \,\bullet \,\vec C = (\vec A\,\bullet \,\vec C) + (\vec B \,\bullet \,\vec C)[/tex]

[tex]\|\vec C\|^{2} = [\vec A \,\bullet \,(\vec A + \vec B)] + [\vec B\,\bullet \,(\vec A + \vec B)][/tex]

[tex]\|\vec C\|^{2} = \vec A \,\bullet \, \vec A + \vec B\,\bullet \vec B + 2\cdot (\vec A \,\bullet \, \vec B)[/tex]

[tex]\|\vec C\|^{2} = \|\vec A\|^{2} + \|\vec B\|^{2} + 2\cdot \|\vec A\|\cdot \|\vec B\|\cdot \cos\theta_{C}[/tex]

Step-by-step explanation:

Let be [tex]\vec A[/tex], [tex]\vec B[/tex] and [tex]\vec C[/tex] the vector of a triangle so that [tex]\vec C = \vec A + \vec B[/tex]. By definition of Dot Product:

[tex]\vec C \,\bullet\,\vec C = (\vec A + \vec B) \,\bullet \vec C[/tex]

[tex]\vec C \,\bullet \,\vec C = (\vec A\,\bullet \,\vec C) + (\vec B \,\bullet \,\vec C)[/tex]

[tex]\|\vec C\|^{2} = [\vec A \,\bullet \,(\vec A + \vec B)] + [\vec B\,\bullet \,(\vec A + \vec B)][/tex]

[tex]\|\vec C\|^{2} = \vec A \,\bullet \, \vec A + \vec B\,\bullet \vec B + 2\cdot (\vec A \,\bullet \, \vec B)[/tex]

[tex]\|\vec C\|^{2} = \|\vec A\|^{2} + \|\vec B\|^{2} + 2\cdot \|\vec A\|\cdot \|\vec B\|\cdot \cos\theta_{C}[/tex]

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