Question

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

equal to –

• A. 30⁰
• B. 60⁰
• C. 75⁰
• D. 90⁰

Answer 1

Answer: (B)

60⁰

Answer 2

The given relation can be written as

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

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

⇒ (a + b)² + 3c(a + b) + 2c² = 3(ab + ac + bc+ c²)

⇒ a² + b² – c² = ab

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

= = $\frac { 1 } { 2 }$

⇒ C = 60⁰.