Question

In a triangle \(ABC\), if \(\angle A=60^\circ\), then \[ \frac{b}{c+a}+\frac{c}{a+b}= \]

• A. \(a+b+c\)
• B. \(0\)
• C. \(1\)
• D. \(abc\)

Hint:

When an angle of a triangle is given, use the cosine rule: \[ a^2=b^2+c^2-2bc\cos A \] to convert side expressions into simpler forms.

Answer 1

The Correct Option is C

Step 1: Start with the given expression.
We need to find the value of \[ \frac{b}{c+a}+\frac{c}{a+b} \]

Step 2: Take the LCM.
\[ \frac{b}{c+a}+\frac{c}{a+b} = \frac{b(a+b)+c(a+c)}{(a+c)(a+b)} \]

Step 3: Simplify the numerator.
\[ b(a+b)+c(a+c)=ab+b^2+ac+c^2 \] \[ =a(b+c)+b^2+c^2 \] So, \[ \frac{b}{c+a}+\frac{c}{a+b} = \frac{a(b+c)+b^2+c^2}{(a+c)(a+b)} \]

Step 4: Simplify the denominator.
\[ (a+c)(a+b)=a^2+ab+ac+bc \] \[ =a^2+a(b+c)+bc \]

Step 5: Use cosine rule.
In triangle \(ABC\), \[ a^2=b^2+c^2-2bc\cos A \] Given, \[ A=60^\circ \] So, \[ a^2=b^2+c^2-2bc\cos 60^\circ \] Since \[ \cos 60^\circ=\frac{1}{2}, \] we get \[ a^2=b^2+c^2-bc \] Therefore, \[ b^2+c^2=a^2+bc \]

Step 6: Substitute in the numerator.
The numerator is \[ a(b+c)+b^2+c^2 \] Using \[ b^2+c^2=a^2+bc, \] we get \[ a(b+c)+b^2+c^2 = a(b+c)+a^2+bc \] \[ =a^2+a(b+c)+bc \] This is exactly the denominator.

Step 7: Final conclusion.
Therefore, \[ \frac{b}{c+a}+\frac{c}{a+b}=1 \] Hence, \[ \boxed{1} \]