If `sin(120^(@)-A)=sin(120^(@)-B),0ltA,Bltpi`, then the values A and B are

Options

A. A = B
B. A = B or `A+B=(pi)/(3)`
C. `A+B=0,A+B=(pi)/(3)`
D. none of these

Answer

To solve the equation \( \sin(120^\circ - A) = \sin(120^\circ - B) \) under the constraints \( 0 < A, B < \pi \), we can follow these steps:

### Step 1: Use the property of sine
We know that if \( \sin x = \sin y \), then:
1. \( x = y + 360^\circ k \) for any integer \( k \)
2. \( x = 180^\circ - y + 360^\circ k \) for any integer \( k \)

In our case, we can set:
\[ 120^\circ - A = 120^\circ - B \quad \text{(Case 1)} \]
or
\[ 120^\circ - A = 180^\circ - (120^\circ - B) \quad \text{(Case 2)} \]

### Step 2: Solve Case 1
From Case 1:
\[ 120^\circ - A = 120^\circ - B \]
This simplifies to:
\[ A = B \]

### Step 3: Solve Case 2
From Case 2:
\[ 120^\circ - A = 180^\circ - 120^\circ + B \]
This simplifies to:
\[ 120^\circ - A = 60^\circ + B \]
Rearranging gives:
\[ A + B = 120^\circ - 60^\circ \]
\[ A + B = 60^\circ \]

### Step 4: Combine results
From the two cases, we have:
1. \( A = B \)
2. \( A + B = 60^\circ \)

Substituting \( B = A \) into the second equation gives:
\[ A + A = 60^\circ \]
\[ 2A = 60^\circ \]
\[ A = 30^\circ \]
Thus, \( B = 30^\circ \) as well.

### Final Answer
The values of \( A \) and \( B \) are:
\[ A = 30^\circ, \quad B = 30^\circ \]