The angles of a triangle ABC are in A.P. and `b:c=sqrt(3):sqrt(2)," find "angleA`.

Answer

To find angle A in triangle ABC where the angles are in arithmetic progression (A.P.) and the ratio of sides b:c is \(\sqrt{3}:\sqrt{2}\), we can follow these steps:

### Step-by-Step Solution:

1. **Define the Angles**:
Let the angles of triangle ABC be:
- Angle A = \(A\)
- Angle B = \(A + D\)
- Angle C = \(A + 2D\)

2. **Sum of Angles in a Triangle**:
The sum of the angles in any triangle is \(180^\circ\). Therefore, we can write:
\[ A + (A + D) + (A + 2D) = 180^\circ \]
Simplifying this, we get:
\[ 3A + 3D = 180^\circ \]
Dividing the entire equation by 3:
\[ A + D = 60^\circ \]

3. **Expressing Angles**:
From the equation \(A + D = 60^\circ\), we can express angle B as:
\[ B = A + D = 60^\circ \]

4. **Using the Ratio of Sides**:
We are given that the ratio of sides \(b:c = \sqrt{3}:\sqrt{2}\). By the Law of Sines, we know:
\[ \frac{b}{c} = \frac{\sin B}{\sin C} \]
Substituting the known values:
\[ \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin(60^\circ)}{\sin C} \]

5. **Calculating \(\sin(60^\circ)\)**:
We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Therefore, substituting this into the equation gives:
\[ \frac{\sqrt{3}}{\sqrt{2}} = \frac{\frac{\sqrt{3}}{2}}{\sin C} \]

6. **Cross-Multiplying**:
Cross-multiplying to solve for \(\sin C\):
\[ \sqrt{3} \cdot \sin C = \frac{\sqrt{3}}{2} \cdot \sqrt{2} \]
Simplifying this:
\[ \sin C = \frac{\frac{\sqrt{3}}{2} \cdot \sqrt{2}}{\sqrt{3}} = \frac{1}{\sqrt{2}} \]

7. **Finding Angle C**:
We know that \(\sin C = \frac{1}{\sqrt{2}}\), which corresponds to:
\[ C = 45^\circ \]

8. **Finding Angle A**:
Now that we have angles B and C, we can find angle A:
\[ A = 180^\circ - B - C = 180^\circ - 60^\circ - 45^\circ = 75^\circ \]

### Final Answer:
Thus, the measure of angle A is:
\[ \boxed{75^\circ} \]