In a `triangleABC," if "C=60^(@)," then "(a)/(b+c)+(b)/(c+a)=`

Options

A. 2
B. 1
C. 4
D. none of these

Answer

To solve the problem step by step, we will analyze the given information and apply the properties of triangles.

### Given:
In triangle \( ABC \), angle \( C = 60^\circ \).

We need to find the value of:
\[ \frac{a}{b+c} + \frac{b}{c+a} \]

### Step 1: Combine the fractions
To combine the fractions, we will find a common denominator:
\[ \frac{a}{b+c} + \frac{b}{c+a} = \frac{a(c+a) + b(b+c)}{(b+c)(c+a)} \]

### Step 2: Expand the numerator
Now, we will expand the numerator:
\[ = \frac{ac + a^2 + b^2 + bc}{(b+c)(c+a)} \]

### Step 3: Use the cosine rule
According to the cosine rule, we have:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Since \( C = 60^\circ \), we know that:
\[ \cos 60^\circ = \frac{1}{2} \]
Thus, we can set up the equation:
\[ \frac{1}{2} = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 4: Rearranging the cosine rule equation
Multiplying both sides by \( 2ab \):
\[ ab = a^2 + b^2 - c^2 \]
Rearranging gives us:
\[ a^2 + b^2 = ab + c^2 \]

### Step 5: Substitute back into the combined fraction
Now, we substitute \( a^2 + b^2 \) into our combined fraction:
\[ = \frac{ac + ab + c^2 + bc}{(b+c)(c+a)} \]

### Step 6: Simplifying the expression
The numerator simplifies to:
\[ = \frac{ac + ab + c^2 + bc}{bc + c^2 + ab + ac} \]

### Step 7: Final simplification
This can be simplified to:
\[ = 1 \]

### Conclusion
Thus, we find that:
\[ \frac{a}{b+c} + \frac{b}{c+a} = 1 \]

### Answer
The final answer is:
\[ 1 \]