Question

$\Large{\mathsf{Challenge:-}}$
In a triangle ABC,

$\large{\mathsf{ \frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c} }}$
Find the measure of angle C.
Where a, b and c are sides.

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Answer 1

Step-by-step explanation:

We have,

$\frac{1}{a + c} + \frac{1}{b + c} = \frac{3}{a + b + c}$

$\implies\frac{a + b + 2c}{(a + c)(b + c)} = \frac{3}{a + b + c}$

$\implies(a + b + 2c)(a + b + c) = 3 (a + c)(b + c)$

$\implies(a + b + c + c)(a + b + c) = 3 (a + c)(b + c)$

$\implies(a + b + c )^{2} + c(a + b + c) = 3(ab + ac + cb + {c}^{2})$

$\implies \: a^{2} + b^{2} + c ^{2} + 2ab + 2bc + 2ca + ca + b c+ c ^{2} = 3ab + 3ac + 3cb + 3{c}^{2}$

$\implies \: a^{2} + b^{2} + 2ab = 3ab + {c}^{2}$

$\implies \: a^{2} + b^{2} - {c}^{2} = 3ab - 2ab$

$\implies \: a^{2} + b^{2} - {c}^{2} =ab$

$\implies \: \frac{a^{2} + b^{2} - {c}^{2} }{ab} =1$

$\implies \: \frac{a^{2} + b^{2} - {c}^{2} }{2ab} = \frac{1}{2}$

$\implies \: \frac{a^{2} + b^{2} - {c}^{2} }{2ab} = \frac{1}{2} = \cos(C)$

$\implies \: \cos(C) = \frac{1}{2}$

$\implies \: C = \frac{\pi}{3}$

Answer 2

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