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

To determine the type of triangle \( \triangle ABC \) given that \( \sin A \sin B = \frac{ab}{c^2} \), we start by considering what implication this equation has on the triangle's properties.

1. In any triangle, \(\sin{A} = \frac{a}{2R}\) and \(\sin{B} = \frac{b}{2R}\), where \( R \) is the circumradius of the triangle.
2. Substituting these into the given equation, we have: \(\left(\frac{a}{2R}\right) \left(\frac{b}{2R}\right) = \frac{ab}{c^2}\). Simplifying, we get: \(\frac{ab}{4R^2} = \frac{ab}{c^2}\).
3. Canceling \(ab\) from both sides, this becomes: \(\frac{1}{4R^2} = \frac{1}{c^2}\) or \(4R^2 = c^2\).
4. This simplifies to: \(R = \frac{c}{2}\). This is a well-known result in the case of a right-angled triangle, where the hypotenuse (longest side) equals the diameter of the circumcircle.
5. Therefore, the equation \( \sin A \sin B = \frac{ab}{c^2} \) implies that \( \triangle ABC \) is a right-angled triangle, with \( c \) being the hypotenuse.

Thus, the correct answer is that the triangle is right angled.