Question

In a \(\triangle ABC\), if \(\sin A \sin B = \frac{ab}{c^2}\), then the triangle is

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

Hint:

Always use sine rule to relate sides and sines in triangle problems.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Use sine rule: \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R\).

Step 2: Detailed Explanation:
Given \(\sin A \sin B = \frac{ab}{c^2}\).
By sine rule, \(a = 2R \sin A\), \(b = 2R \sin B\), \(c = 2R \sin C\).
RHS = \(\frac{(2R \sin A)(2R \sin B)}{(2R \sin C)^2} = \frac{4R^2 \sin A \sin B}{4R^2 \sin^2 C} = \frac{\sin A \sin B}{\sin^2 C}\).
So equation becomes \(\sin A \sin B = \frac{\sin A \sin B}{\sin^2 C}\)
\(\implies \sin^2 C = 1 \implies \sin C = 1 \implies C = 90^\circ\).

Step 3: Final Answer:
Triangle is right angled. Option (C).