Question

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

Answer 1

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

`\vec C \,\bullet\,\vec C = (\vec A + \vec B) \,\bullet \vec C`

`\vec C \,\bullet \,\vec C = (\vec A\,\bullet \,\vec C) + (\vec B \,\bullet \,\vec C)`

`\|\vec C\|^(2) = [\vec A \,\bullet \,(\vec A + \vec B)] + [\vec B\,\bullet \,(\vec A + \vec B)]`

`\|\vec C\|^(2) = \vec A \,\bullet \, \vec A + \vec B\,\bullet \vec B + 2\cdot (\vec A \,\bullet \, \vec B)`

`\|\vec C\|^(2) = \|\vec A\|^(2) + \|\vec B\|^(2) + 2\cdot \|\vec A\|\cdot \|\vec B\|\cdot \cos\theta_(C)`

Explanation:

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

`\vec C \,\bullet\,\vec C = (\vec A + \vec B) \,\bullet \vec C`

`\vec C \,\bullet \,\vec C = (\vec A\,\bullet \,\vec C) + (\vec B \,\bullet \,\vec C)`

`\|\vec C\|^(2) = [\vec A \,\bullet \,(\vec A + \vec B)] + [\vec B\,\bullet \,(\vec A + \vec B)]`

`\|\vec C\|^(2) = \vec A \,\bullet \, \vec A + \vec B\,\bullet \vec B + 2\cdot (\vec A \,\bullet \, \vec B)`

`\|\vec C\|^(2) = \|\vec A\|^(2) + \|\vec B\|^(2) + 2\cdot \|\vec A\|\cdot \|\vec B\|\cdot \cos\theta_(C)`