Question

In a triangle ABC, if ${\displaystyle \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}}$ then angle C is equal to ______

Options

• 30°

• 60°

• 90°

• 120°

Answer 1

In a triangle ABC, if ${\displaystyle \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}}$ then angle C is equal to 60°.

Explanation:

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

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

⇒ (a + b + 2c) (a + b + c) = 3(a + c) (b + c)

⇒ a² + b² - c² = ab

∴ cos C = ${\displaystyle \frac{a^2+b^2-c^2}{2ab}}$

= ${\displaystyle \frac{4+3-a^2}{4\sqrt{3}}}$

= ${\displaystyle \frac{7-a^2}{4\sqrt{3}}}$

= cos 60°

⇒ C = 60°