In a triangle ABC, if `sin A sin B= (ab)/(c^(2))`, then the triangle is :

Options

A. (a)Equilateral
B. (b)Isosceles
C. (c)Right angled
D. (d)Obtuse angled

Explanation

To solve the problem, we need to analyze the given equation in the context of triangle properties. The equation given is:

\[ \sin A \sin B = \frac{ab}{c^2} \]

where \( A \), \( B \), and \( C \) are the angles of triangle \( ABC \), and \( a \), \( b \), and \( c \) are the sides opposite to these angles respectively.

### Step-by-Step Solution:

1. **Start with the given equation**:
\[ \sin A \sin B = \frac{ab}{c^2} \]

2. **Use the Law of Sines**:
According to the Law of Sines, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]
From this, we can express \( a \) and \( b \) in terms of \( c \) and the angles:
\[ a = \frac{c \sin A}{\sin C}, \quad b = \frac{c \sin B}{\sin C} \]

3. **Substitute \( a \) and \( b \) into the equation**:
Substitute these expressions into the original equation:
\[ \sin A \sin B = \frac{\left(\frac{c \sin A}{\sin C}\right)\left(\frac{c \sin B}{\sin C}\right)}{c^2} \]
Simplifying this gives:
\[ \sin A \sin B = \frac{c^2 \sin A \sin B}{c^2 \sin^2 C} \]

4. **Cancel \( \sin A \sin B \) and \( c^2 \)**:
If \( \sin A \sin B \neq 0 \), we can divide both sides by \( \sin A \sin B \):
\[ 1 = \frac{1}{\sin^2 C} \]
This implies:
\[ \sin^2 C = 1 \]

5. **Find angle \( C \)**:
Taking the square root of both sides, we find:
\[ \sin C = 1 \]
This means:
\[ C = 90^\circ \]

6. **Conclusion**:
Since angle \( C \) is \( 90^\circ \), triangle \( ABC \) is a right-angled triangle.

### Final Answer:
The triangle is a **right-angled triangle**.