If `sinA=sin^2B and 2cos^2A=3cos^2B` then the triangle ABC is

Options

A. right angled
B. obtuse angled
C. ospsceles
D. equilateral

Explanation

To solve the problem, we need to analyze the given equations and find the angles of triangle ABC based on the conditions provided.

### Step-by-Step Solution:

1. **Given Equations**:
- \( \sin A = \sin^2 B \)
- \( 2 \cos^2 A = 3 \cos^2 B \)

2. **Express Cosine in Terms of Sine**:
Using the identity \( \cos^2 \theta = 1 - \sin^2 \theta \):
- From \( 2 \cos^2 A = 3 \cos^2 B \), we can rewrite it as:
\[ 2(1 - \sin^2 A) = 3(1 - \sin^2 B) \]
- This simplifies to:
\[ 2 - 2 \sin^2 A = 3 - 3 \sin^2 B \]
- Rearranging gives:
\[ 2 \sin^2 A - 3 \sin^2 B + 1 = 0 \]

3. **Substituting \( \sin A \)**:
Since \( \sin A = \sin^2 B \), we can substitute this into the equation:
- Let \( x = \sin B \). Then \( \sin A = x^2 \).
- Substitute \( \sin A \) into the equation:
\[ 2(x^2)^2 - 3x^2 + 1 = 0 \]
- This simplifies to:
\[ 2x^4 - 3x^2 + 1 = 0 \]

4. **Letting \( y = x^2 \)**:
- We can let \( y = x^2 \), then the equation becomes:
\[ 2y^2 - 3y + 1 = 0 \]

5. **Solving the Quadratic Equation**:
- Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
- This gives us:
\[ y = 1 \quad \text{or} \quad y = \frac{1}{2} \]

6. **Finding \( \sin B \)**:
- Since \( y = \sin^2 B \):
- If \( y = 1 \), then \( \sin B = 1 \) which implies \( B = 90^\circ \) (not possible in a triangle).
- If \( y = \frac{1}{2} \), then \( \sin B = \frac{1}{\sqrt{2}} \) which implies \( B = 45^\circ \).

7. **Finding \( \sin A \)**:
- From \( \sin A = \sin^2 B \):
\[ \sin A = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
- This implies \( A = 30^\circ \).

8. **Finding Angle C**:
- The sum of angles in a triangle is \( 180^\circ \):
\[ C = 180^\circ - A - B = 180^\circ - 30^\circ - 45^\circ = 105^\circ \]

### Conclusion:
The angles of triangle ABC are:
- \( A = 30^\circ \)
- \( B = 45^\circ \)
- \( C = 105^\circ \)

Since one angle is greater than \( 90^\circ \), triangle ABC is an **obtuse triangle**.