A triangle `ABC` is such that `sin(2A+B) =1/2` and If `A, B` and `C` are in `A.P.,` then find the value of `A` and `C`

Answer

To solve the problem, we need to find the values of angles \( A \) and \( C \) in triangle \( ABC \) given that \( \sin(2A + B) = \frac{1}{2} \) and that \( A, B, C \) are in Arithmetic Progression (A.P.).

### Step 1: Set up the equations
Since \( A, B, C \) are angles of a triangle, we know:
\[ A + B + C = 180^\circ \tag{1} \]
Also, since \( A, B, C \) are in A.P., we can express this as:
\[ 2B = A + C \tag{2} \]

### Step 2: Substitute equation (2) into equation (1)
From equation (2), we can express \( C \) in terms of \( A \) and \( B \):
\[ C = 2B - A \]
Substituting this into equation (1):
\[ A + B + (2B - A) = 180^\circ \]
This simplifies to:
\[ 3B = 180^\circ \]
Thus, we find:
\[ B = 60^\circ \tag{3} \]

### Step 3: Use the value of \( B \) in equation (2)
Now that we have \( B \), we can substitute it back into equation (2):
\[ 2(60^\circ) = A + C \]
This gives:
\[ 120^\circ = A + C \tag{4} \]

### Step 4: Solve the sine equation
Next, we use the sine equation given in the problem:
\[ \sin(2A + B) = \frac{1}{2} \]
Using the value of \( B \) from equation (3):
\[ \sin(2A + 60^\circ) = \frac{1}{2} \]
The angles for which sine is \( \frac{1}{2} \) are:
\[ 2A + 60^\circ = 30^\circ \quad \text{or} \quad 2A + 60^\circ = 150^\circ \]

### Step 5: Solve for \( A \)
**Case 1:**
\[ 2A + 60^\circ = 30^\circ \]
\[ 2A = 30^\circ - 60^\circ = -30^\circ \quad \Rightarrow \quad A = -15^\circ \quad \text{(not valid)} \]

**Case 2:**
\[ 2A + 60^\circ = 150^\circ \]
\[ 2A = 150^\circ - 60^\circ = 90^\circ \quad \Rightarrow \quad A = 45^\circ \tag{5} \]

### Step 6: Find \( C \)
Using equation (4) to find \( C \):
\[ A + C = 120^\circ \]
Substituting \( A = 45^\circ \):
\[ 45^\circ + C = 120^\circ \]
\[ C = 120^\circ - 45^\circ = 75^\circ \tag{6} \]

### Final Values
Thus, the values of angles \( A \) and \( C \) are:
\[ A = 45^\circ, \quad C = 75^\circ \]