Math, asked by Revanthkodukula9849, 1 year ago

In ΔABC, show that \frac{cos\ A}{a} + \frac{cos\ B}{b} + \frac{cos\ C}{c} = \frac{a^{2}\ +\ b^{2}\ +\ c^{2}}{2abc}.

Answers

Answered by MaheswariS
0

Answer:


Step-by-step explanation:


Formula used:


Cosine formulae:

1.a² = b² + c² - 2bc cosA

2.b² = c² + a² - 2ca cosB

3.c² = a² + b² - 2ab cosC


\frac{cosA}{a} + \frac{cosB}{b} + \frac{cosC}{c}

= \frac{bc\:cosA}{abc}+\frac{ca\:cosB}{abc}+\frac{ab\:cosC}{abc}

=\frac{\frac{1}{2}({b}^{2}+{c}^{2}-{a}^{2})+\frac{1}{2}( {c}^{2} +{a}^{2}-{b}^{2})+\frac{1}{2}({a}^{2}+{b}^{2}-{c}^{2})}{abc}

=\frac{\frac{1}{2}({b}^{2}+{c}^{2}-{a}^{2}+{c}^{2} +{a}^{2}-{b}^{2}+{a}^{2}+{b}^{2}-{c}^{2})}{abc}

=\frac{\frac{1}{2}({a}^{2}+{b}^{2} + {c}^{2})}{abc}

=\frac{{a}^{2}+{b}^{2}+{c}^{2}}{2abc}


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