In a triangle $A B C$ , using vectors, prove that $c^{2} = a^{2} + b^{2} - 2 a b cos C$

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Solution

As we have $A B C$ is a triangle hence $\to a + \to b + \to c = \to 0$
So, $\to a + \to b = - \to c$
Squaring on both sides,
$(\to a + \to b) (\to a + \to b) = (- \to c) (- \to c)$
$b^{2} + a^{2} + 2 a b cos (π - C) = c^{2}$ [Using dot product ]
$b^{2} + a^{2} + 2 a b (- cos C) = c^{2}$
$c^{2} = a^{2} + b^{2} - 2 a b cos C$

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