Question

In a triangle ABC, if `1/(a + c) + 1/(b + c) = 3/(a + b + c)` then angle C is equal to ______

विकल्प

• 30°

• 60°

• 90°

• 120°

Answer 1

In a triangle ABC, if `1/(a + c) + 1/(b + c) = 3/(a + b + c)` then angle C is equal to 60°.

Explanation:

`1/(a + c) + 1/(b + c) = 3/(a + b + c)`

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

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

⇒ a² + b² - c² = ab

∴ cos C = `(a^2 + b^2 - c^2)/(2ab)`

= `(4 + 3 - a^2)/(4sqrt3)`

= `(7 - a^2)/(4sqrt3)`

= cos 60°

⇒ C = 60°