Question

In \( \triangle ABC \), if \( \sin^2 B = \sin A \) and \( 2\cos^2 A = 3\cos^2 B \), then the triangle is:

• A. acute angled
• B. obtuse angled
• C. right angled
• D. equilateral

Hint:

Use trigonometric identities and triangle angle sum rule to verify angle types. When one angle exceeds \( 90^\circ \), the triangle is obtuse.

Answer 1

The Correct Option is B

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.