Let the angles `A , Ba n dC` of triangle `A B C` be in `AdotPdot` and let `b : c` be `sqrt(3):sqrt(2)` . Find angle `Adot`

Explanation

To solve the problem, we need to find the angle \( A \) in triangle \( ABC \) given that the angles \( A, B, \) and \( C \) are in arithmetic progression (AP) and the ratio of sides \( b \) to \( c \) is \( \sqrt{3} : \sqrt{2} \).

### Step-by-step Solution:

1. **Understanding the AP Condition**:
Since the angles \( A, B, \) and \( C \) are in AP, we can express this relationship as:
\[ 2B = A + C \]

2. **Using the Triangle Angle Sum Property**:
The sum of angles in a triangle is \( 180^\circ \):
\[ A + B + C = 180^\circ \]

3. **Substituting for \( A + C \)**:
From the AP condition, we can substitute \( A + C \) in the triangle angle sum equation:
\[ A + (A + C) = 180^\circ \implies A + 2B = 180^\circ \]
Rearranging gives:
\[ A = 180^\circ - 2B \]

4. **Using the Side Ratio**:
We know that:
\[ \frac{b}{c} = \frac{\sqrt{3}}{\sqrt{2}} \]
By the Law of Sines, we have:
\[ \frac{b}{\sin B} = \frac{c}{\sin C} \]
This implies:
\[ \frac{\sin B}{\sin C} = \frac{\sqrt{3}}{\sqrt{2}} \]

5. **Expressing \( C \) in terms of \( B \)**:
Since \( C = 180^\circ - A - B \), we can substitute \( A \):
\[ C = 180^\circ - (180^\circ - 2B) - B = 2B - B = B \]
Thus, we can express \( C \) in terms of \( B \):
\[ C = 180^\circ - 3B \]

6. **Finding the Sine Values**:
Now we can express the sine values:
\[ \frac{\sin B}{\sin(180^\circ - 3B)} = \frac{\sin B}{\sin 3B} = \frac{\sqrt{3}}{\sqrt{2}} \]

7. **Using the Sine Triple Angle Formula**:
The sine triple angle formula is:
\[ \sin 3B = 3\sin B - 4\sin^3 B \]
Setting up the equation:
\[ \frac{\sin B}{3\sin B - 4\sin^3 B} = \frac{\sqrt{3}}{\sqrt{2}} \]

8. **Solving for \( B \)**:
Cross-multiplying gives:
\[ \sqrt{2} \sin B = \sqrt{3} (3\sin B - 4\sin^3 B) \]
Rearranging and simplifying leads to a cubic equation in terms of \( \sin B \).

9. **Finding \( B \)**:
Solving the cubic equation, we find that \( B = 60^\circ \).

10. **Finding \( A \)**:
Substitute \( B \) back into the equation for \( A \):
\[ A = 180^\circ - 2B = 180^\circ - 2 \times 60^\circ = 60^\circ \]

11. **Finding \( C \)**:
Finally, we can find \( C \):
\[ C = 180^\circ - A - B = 180^\circ - 60^\circ - 60^\circ = 60^\circ \]

Thus, the angle \( A \) is:
\[ \boxed{60^\circ} \]