If `A=60^@, " then " b/(c+a)+c/(a+b) =`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the problem, we need to find the value of \( \frac{b}{c+a} + \frac{c}{a+b} \) given that \( A = 60^\circ \).

### Step 1: Use the Law of Cosines
We can use the Law of Cosines to relate the sides of the triangle to the angles. The Law of Cosines states that:
\[ A^2 = B^2 + C^2 - 2BC \cos A \]
Given \( A = 60^\circ \), we know that \( \cos 60^\circ = \frac{1}{2} \). Thus, we can rewrite the equation as:
\[ A^2 = B^2 + C^2 - BC \]

### Step 2: Rearranging the Equation
Rearranging the equation gives us:
\[ A^2 + BC = B^2 + C^2 \]

### Step 3: Find \( \frac{b}{c+a} + \frac{c}{a+b} \)
Now, we need to find \( \frac{b}{c+a} + \frac{c}{a+b} \). We can combine these fractions:
\[ \frac{b}{c+a} + \frac{c}{a+b} = \frac{b(a+b) + c(c+a)}{(c+a)(a+b)} \]

### Step 4: Simplifying the Numerator
Expanding the numerator:
\[ b(a+b) + c(c+a) = ab + b^2 + c^2 + ac \]
So we have:
\[ \frac{ab + b^2 + c^2 + ac}{(c+a)(a+b)} \]

### Step 5: Substitute \( A^2 \)
From our earlier rearrangement, we know that \( b^2 + c^2 = A^2 + BC \). Substituting this into our expression gives:
\[ \frac{ab + A^2 + BC + ac}{(c+a)(a+b)} \]

### Step 6: Final Expression
We can simplify this further, but we need to evaluate it based on the values of \( a, b, c \) and \( A \). However, without specific values for \( a, b, c \), we can conclude that:
\[ \frac{b}{c+a} + \frac{c}{a+b} = \frac{ab + A^2 + BC + ac}{(c+a)(a+b)} \]

### Final Answer
Thus, the expression \( \frac{b}{c+a} + \frac{c}{a+b} \) simplifies to the above fraction, depending on the values of \( a, b, c \).

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