If `A,B,C` are the angles of a triangle, the system of equations
`(sinA)x+y+z=cosA`, `x+(sinB)y+z=cosB`, `x+y+(sinC)z=1-cosC` has

Options

A. No solution
B. Unique solution
C. Infinitely many solutions
D. Finitely many solutions

Explanation

To determine the type of solution for the given system of equations involving the angles of a triangle, we will calculate the determinant of the coefficient matrix. The system of equations is:

1. \( (\sin A)x + y + z = \cos A \)
2. \( x + (\sin B)y + z = \cos B \)
3. \( x + y + (\sin C)z = 1 - \cos C \)

### Step 1: Write the coefficient matrix and the constant matrix

The coefficient matrix \( M \) and the constant matrix \( C \) can be represented as follows:

\[ M = \begin{pmatrix} \sin A & 1 & 1 \\ 1 & \sin B & 1 \\ 1 & 1 & \sin C \end{pmatrix}, \quad C = \begin{pmatrix} \cos A \\ \cos B \\ 1 - \cos C \end{pmatrix} \]

### Step 2: Calculate the determinant of the coefficient matrix

We will calculate the determinant of matrix \( M \):

\[ \text{det}(M) = \begin{vmatrix} \sin A & 1 & 1 \\ 1 & \sin B & 1 \\ 1 & 1 & \sin C \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix, we have:

\[ \text{det}(M) = \sin A \begin{vmatrix} \sin B & 1 \\ 1 & \sin C \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 1 & \sin C \end{vmatrix} + 1 \begin{vmatrix} 1 & \sin B \\ 1 & 1 \end{vmatrix} \]

Calculating the 2x2 determinants:

1. \( \begin{vmatrix} \sin B & 1 \\ 1 & \sin C \end{vmatrix} = \sin B \sin C - 1 \)

2. \( \begin{vmatrix} 1 & 1 \\ 1 & \sin C \end{vmatrix} = \sin C - 1 \)

3. \( \begin{vmatrix} 1 & \sin B \\ 1 & 1 \end{vmatrix} = 1 - \sin B \)

Substituting these back into the determinant formula:

\[ \text{det}(M) = \sin A (\sin B \sin C - 1) - (\sin C - 1) + (1 - \sin B) \]

Simplifying this gives:

\[ \text{det}(M) = \sin A \sin B \sin C - \sin A - \sin B - \sin C + 2 \]

### Step 3: Analyze the determinant

Since \( A, B, C \) are angles of a triangle, they lie in the range \( (0, \pi) \). Therefore, \( \sin A, \sin B, \sin C > 0 \).

We can conclude that:

- \( \sin A \sin B \sin C > 0 \)
- The expression \( -\sin A - \sin B - \sin C + 2 \) can be analyzed. Since \( \sin A, \sin B, \sin C \) are positive and less than or equal to 1, the sum \( -\sin A - \sin B - \sin C \) can be at least -3, but the +2 ensures that the determinant is positive.

Thus, we conclude that:

\[ \text{det}(M) > 0 \]

### Step 4: Conclusion about the system of equations

Since the determinant of the coefficient matrix is greater than zero, the system of equations has a unique solution.