Step 1: Use identity:
We are given:
\[
\sin^2 B = \sin A
\text{and}
2\cos^2 A = 3\cos^2 B
\]
Step 2: Let us assume \( A = x \), \( B = y \)
Then,
\[
\sin^2 y = \sin x \tag{1}
\]
\[
2\cos^2 x = 3\cos^2 y \tag{2}
\]
Step 3: Use identity \( \cos^2 \theta = 1 - \sin^2 \theta \)
Substitute into equation (2):
\[
2(1 - \sin^2 x) = 3(1 - \sin^2 y)
\Rightarrow 2 - 2\sin^2 x = 3 - 3\sin^2 y
\]
\[
\Rightarrow -2\sin^2 x + 2 = -3\sin^2 y + 3
\Rightarrow 3\sin^2 y - 2\sin^2 x = 1
\]
From equation (1): \( \sin^2 y = \sin x \Rightarrow \sin^2 y = \sqrt{\sin^2 x} = \sin x \), assuming angles are in first quadrant.
Now substitute:
\[
3\sin x - 2\sin^2 x = 1
\Rightarrow 2\sin^2 x - 3\sin x + 1 = 0
\]
Step 4: Solve the quadratic in \( \sin x \)
\[
2\sin^2 x - 3\sin x + 1 = 0
\Rightarrow \sin x = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4}
\Rightarrow \sin x = 1 \text{ or } \frac{1}{2}
\]
Step 5: Analyze both values
- If \( \sin x = 1 \Rightarrow x = 90^\circ \)
- If \( \sin x = \frac{1}{2} \Rightarrow x = 30^\circ \)
If \( A = 90^\circ \), triangle would be right-angled.
But test both in equation (1):
If \( \sin x = 1 \), then \( \sin^2 y = 1 \Rightarrow \sin y = 1 \Rightarrow y = 90^\circ \)
That contradicts triangle sum property \( A + B + C = 180^\circ \), because two right angles are not possible.
So use \( \sin x = \frac{1}{2} \Rightarrow x = 30^\circ \Rightarrow \sin^2 y = \frac{1}{2} \Rightarrow \sin y = \frac{1}{\sqrt{2}} \Rightarrow y = 45^\circ \)
Then \( C = 180^\circ - (30^\circ + 45^\circ) = 105^\circ \)
Step 6: Conclusion
One angle is \(>90^\circ \Rightarrow \triangle ABC \) is an obtuse-angled triangle.