Question

In $\triangle ABC$, if $\sin A = \sin^{2} B$ and $2 \cos^{2} A = 3 \cos^{2} B$ then the triangle ABC is

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

Hint:

$\sin \theta = 1$ implies $\theta = 90^{\circ}$, which immediately identifies a right-angled triangle.

Answer 1

The Correct Option is D

Step 1: Concept
Use the identity $\cos^{2} \theta = 1 - \sin^{2} \theta$ to create a consistent equation for one angle.

Step 2: Meaning
Substitute $\sin^{2} B = \sin A$ into the second equation: $2 \cos^{2} A = 3(1 - \sin^{2} B)$.

Step 3: Analysis
$2(1 - \sin^{2} A) = 3(1 - \sin A) \implies 2 - 2 \sin^{2} A = 3 - 3 \sin A \implies 2 \sin^{2} A - 3 \sin A + 1 = 0$. Solving gives $\sin A = 1$ or $\sin A = 1/2$.

Step 4: Conclusion
If $\sin A = 1$, then $A = 90^{\circ}$. This indicates that the triangle is right-angled. Final Answer: (D)