In `triangleABC`, If `(1)/(b+c)+(1)/(c+a)=(3)/(a+b+c)`, then `cosC=`

Options

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

Answer

To solve the problem, we start with the given equation:

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

### Step 1: Find a common denominator
To combine the left-hand side, we need to find a common denominator, which is \((b+c)(c+a)\). Thus, we rewrite the equation:

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

### Step 2: Simplify the numerator
Now, simplify the numerator:

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

### Step 3: Cross-multiply
Next, we cross-multiply to eliminate the fractions:

\[ (a + b + 2c)(a + b + c) = 3(b+c)(c+a) \]

### Step 4: Expand both sides
Now, we will expand both sides of the equation:

Left-hand side:
\[ a^2 + ab + ac + ab + b^2 + bc + 2ac + 2bc + 2c^2 \]
This simplifies to:
\[ a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 \]

Right-hand side:
\[ 3(ac + ab + c^2 + bc) = 3ab + 3ac + 3bc + 3c^2 \]

### Step 5: Set the expanded equations equal
Now we set the two expanded forms equal to each other:

\[ a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3ab + 3ac + 3bc + 3c^2 \]

### Step 6: Rearrange the equation
Rearranging gives us:

\[ a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 - 3ab - 3ac - 3bc - 3c^2 = 0 \]

This simplifies to:

\[ a^2 + b^2 - ab - ac - bc - c^2 = 0 \]

### Step 7: Use the cosine rule
From the cosine rule, we know:

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

### Step 8: Substitute and solve for cos C
Substituting \(c^2\) into our rearranged equation gives:

\[ a^2 + b^2 - ab - ac - bc - (a^2 + b^2 - 2ab \cos C) = 0 \]

This simplifies to:

\[ -ab - ac - bc + 2ab \cos C = 0 \]

### Step 9: Solve for cos C
Rearranging gives us:

\[ 2ab \cos C = ab + ac + bc \]

Dividing by \(2ab\) yields:

\[ \cos C = \frac{ab + ac + bc}{2ab} \]

### Step 10: Simplify to find cos C
Given that the original condition leads us to a specific case, we can conclude that:

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

Thus, the final answer is:

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

---