If a,b,c are the sides of triangle ABC satisfying `log (1 +c/a)+log a - log b = log 2.` Also `a(1 - x^2) + 2 b x+c(1 + x^2) = 0` has two equal roots. Find the value of `sin A + sin B + sin C.`

Options

A. `(5)/(2)`
B. `(12)/(5)`
C. `(8)/(3)`
D. 2

Answer

To solve the problem step-by-step, we will break down the given equations and find the required value of \( \sin A + \sin B + \sin C \).

### Step 1: Simplifying the logarithmic equation
We start with the equation:
\[ \log(1 + \frac{c}{a}) + \log a - \log b = \log 2 \]
Using the properties of logarithms, we can combine the terms:
\[ \log\left(\frac{a(1 + \frac{c}{a})}{b}\right) = \log 2 \]
This simplifies to:
\[ \log\left(\frac{a + c}{b}\right) = \log 2 \]
Taking the antilogarithm gives us:
\[ \frac{a + c}{b} = 2 \]
Thus, we can rearrange this to find:
\[ a + c = 2b \tag{1} \]

### Step 2: Analyzing the quadratic equation
Next, we consider the quadratic equation:
\[ a(1 - x^2) + 2bx + c(1 + x^2) = 0 \]
Rearranging gives:
\[ (a + c)x^2 + 2bx + a = 0 \]
For this quadratic to have two equal roots, the discriminant must be zero:
\[ (2b)^2 - 4(a + c)(a) = 0 \]
This simplifies to:
\[ 4b^2 - 4a(a + c) = 0 \]
Dividing by 4:
\[ b^2 = a(a + c) \tag{2} \]

### Step 3: Substituting the first equation into the second
From equation (1), we substitute \( c = 2b - a \) into equation (2):
\[ b^2 = a(a + (2b - a)) = a(2b) \]
This simplifies to:
\[ b^2 = 2ab \]
Assuming \( b \neq 0 \), we can divide by \( b \):
\[ b = 2a \tag{3} \]

### Step 4: Finding the value of \( c \)
Substituting \( b = 2a \) back into equation (1):
\[ a + c = 2(2a) = 4a \]
Thus:
\[ c = 4a - a = 3a \tag{4} \]

### Step 5: Finding the sides in terms of a
Now we have:
- \( a = a \)
- \( b = 2a \)
- \( c = 3a \)

### Step 6: Using the sine rule
We know that in a triangle:
\[ \frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} \]
Using the values we found:
\[ \frac{\sin A}{a} = \frac{\sin B}{2a} = \frac{\sin C}{3a} \]
This implies:
\[ \sin A = k \cdot a, \quad \sin B = k \cdot 2a, \quad \sin C = k \cdot 3a \]
for some constant \( k \).

### Step 7: Finding the value of \( k \)
Using the fact that \( A + B + C = 180^\circ \) and knowing that \( C \) is the largest angle (since \( c > b > a \)), we can conclude that \( C \) is \( 90^\circ \). Hence:
\[ \sin C = 1 \]
This gives us:
\[ k \cdot 3a = 1 \implies k = \frac{1}{3a} \]

### Step 8: Finding \( \sin A \) and \( \sin B \)
Now substituting \( k \) back:
\[ \sin A = \frac{1}{3a} \cdot a = \frac{1}{3}, \quad \sin B = \frac{1}{3a} \cdot 2a = \frac{2}{3} \]

### Step 9: Summing the sines
Finally, we find:
\[ \sin A + \sin B + \sin C = \frac{1}{3} + \frac{2}{3} + 1 = 1 + 1 = 2 \]

### Final Answer
Thus, the value of \( \sin A + \sin B + \sin C \) is:
\[ \boxed{2} \]