In `DeltaABC`, if `(1)/(b+c)+(1)/(c+a)=(3)/(a+b+c)`, then C is equal to

Options

A. `90^(@)`
B. `60^(@)`
C. `45^(@)`
D. `30^(@)`

Explanation

To solve the problem, we start with the given equation for triangle \( \Delta ABC \):

\[ \frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c} \]

### Step 1: Cross-Multiply
We can cross-multiply to eliminate the fractions. This gives us:

\[ (a+b+c) \left( \frac{1}{b+c} + \frac{1}{c+a} \right) = 3 \]

Expanding the left-hand side:

\[ (a+b+c) \cdot \frac{1}{b+c} + (a+b+c) \cdot \frac{1}{c+a} = 3 \]

This simplifies to:

\[ \frac{a+b+c}{b+c} + \frac{a+b+c}{c+a} = 3 \]

### Step 2: Simplify Each Term
Now we can simplify each term:

\[ \frac{a}{b+c} + \frac{b+c}{b+c} + \frac{b}{c+a} + \frac{c+a}{c+a} = 3 \]

This simplifies to:

\[ \frac{a}{b+c} + 1 + \frac{b}{c+a} + 1 = 3 \]

So we have:

\[ \frac{a}{b+c} + \frac{b}{c+a} + 2 = 3 \]

### Step 3: Rearranging the Equation
Rearranging gives us:

\[ \frac{a}{b+c} + \frac{b}{c+a} = 1 \]

### Step 4: Finding a Common Denominator
To combine the fractions, we find a common denominator:

\[ \frac{a(c+a) + b(b+c)}{(b+c)(c+a)} = 1 \]

This leads to:

\[ a(c+a) + b(b+c) = (b+c)(c+a) \]

### Step 5: Expanding Both Sides
Expanding both sides gives us:

\[ ac + a^2 + b^2 + bc = bc + c^2 + ab + ac \]

### Step 6: Canceling Terms
We can cancel \( ac \) and \( bc \) from both sides:

\[ a^2 + b^2 = c^2 + ab \]

### Step 7: Rearranging for the Cosine Rule
Rearranging gives us:

\[ a^2 + b^2 - c^2 = ab \]

### Step 8: Applying the Cosine Rule
By the cosine rule, we know:

\[ c^2 = a^2 + b^2 - 2ab \cos C \]

Setting the two equations equal gives:

\[ ab = 2ab \cos C \]

### Step 9: Solving for Cosine
Dividing both sides by \( ab \) (assuming \( ab \neq 0 \)):

\[ 1 = 2 \cos C \]

Thus,

\[ \cos C = \frac{1}{2} \]

### Step 10: Finding Angle C
The angle \( C \) corresponding to \( \cos C = \frac{1}{2} \) is:

\[ C = 60^\circ \]

### Final Answer
Therefore, the angle \( C \) is:

\[ \boxed{60^\circ} \]