The angles A, B and C of a triangle ABC are in arithmetic progression. If `2b^(2)=3c^(2)` then the angle A is :

Options

A. `15^(@)`
B. `60^(@)`
C. `75^(@)`
D. `90^(@)`

Answer

To solve the problem, we need to find the angle A of triangle ABC given that the angles are in arithmetic progression and that \(2b^2 = 3c^2\).

### Step-by-Step Solution:

1. **Define the Angles**:
Since the angles A, B, and C are in arithmetic progression, we can express them as:
- Let \(A = a\)
- Let \(B = a + d\)
- Let \(C = a + 2d\)

2. **Sum of Angles in a Triangle**:
The sum of the angles in any triangle is 180 degrees. Therefore, we can write:
\[ A + B + C = 180^\circ \]
Substituting the expressions for A, B, and C:
\[ a + (a + d) + (a + 2d) = 180^\circ \]
Simplifying this gives:
\[ 3a + 3d = 180^\circ \]
Dividing through by 3:
\[ a + d = 60^\circ \]

3. **Finding Angle B**:
From the equation \(a + d = 60^\circ\), we can find angle B:
\[ B = a + d = 60^\circ \]

4. **Using the Given Condition**:
We are given that \(2b^2 = 3c^2\). This can be rearranged to:
\[ \frac{b^2}{c^2} = \frac{3}{2} \]
Taking the square root of both sides gives:
\[ \frac{b}{c} = \sqrt{\frac{3}{2}} = \frac{\sqrt{3}}{\sqrt{2}} \]

5. **Applying the Sine Rule**:
By the sine rule, we know:
\[ \frac{b}{c} = \frac{\sin B}{\sin C} \]
Substituting the known value of angle B:
\[ \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^\circ}{\sin C} \]
Since \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), we can substitute this in:
\[ \frac{\sqrt{3}}{\sqrt{2}} = \frac{\frac{\sqrt{3}}{2}}{\sin C} \]
Cross-multiplying gives:
\[ \sqrt{3} \cdot \sin C = \frac{\sqrt{3}}{2} \cdot \sqrt{2} \]
Simplifying this yields:
\[ \sin C = \frac{1}{\sqrt{2}} \Rightarrow C = 45^\circ \]

6. **Finding Angle A**:
Now we can find angle A using the angle sum property:
\[ A + B + C = 180^\circ \]
Substituting the known values:
\[ A + 60^\circ + 45^\circ = 180^\circ \]
Simplifying gives:
\[ A + 105^\circ = 180^\circ \]
Therefore:
\[ A = 180^\circ - 105^\circ = 75^\circ \]

### Final Answer:
Thus, the angle A is \(75^\circ\).