In `DeltaABC`, If `1/(a+b)+1/(b+c)=3/(a+b+c)`, then `angleB=?`

Options

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

Answer

To solve the problem, we need to find the angle \( B \) in triangle \( ABC \) given the equation:

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

### Step 1: Rewrite the given equation

Start by rewriting the equation:

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

### Step 2: Find a common denominator

The left-hand side can be combined using a common denominator:

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

This simplifies to:

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

### Step 3: Cross-multiply

Cross-multiply to eliminate the fractions:

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

### Step 4: Expand both sides

Now, expand both sides:

Left-hand side:

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

This simplifies to:

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

Right-hand side:

\[ 3(ab + ac + b^2 + bc) = 3ab + 3ac + 3b^2 + 3bc \]

### Step 5: Set the expanded forms equal

Set the two expanded forms equal to each other:

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

### Step 6: Rearrange the equation

Rearranging gives:

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

This simplifies to:

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

### Step 7: Use the Cosine Rule

From the cosine rule in triangle \( ABC \):

\[ a^2 + c^2 - b^2 = 2ac \cos B \]

Substituting gives:

\[ 2ac \cos B = a^2 + c^2 - b^2 \]

### Step 8: Divide by \( 2ac \)

Divide the entire equation by \( 2ac \):

\[ \cos B = \frac{a^2 + c^2 - b^2}{2ac} \]

### Step 9: Substitute and simplify

From our rearranged equation, we have:

\[ \frac{a^2 + c^2 - b^2}{2ac} = \frac{1}{2} \]

This implies:

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

### Step 10: Find angle \( B \)

The angle \( B \) for which \( \cos B = \frac{1}{2} \) is:

\[ B = 60^\circ \]

### Final Answer

Thus, the angle \( B \) is:

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