In `DeltaABC`, with the usual notations, if `sinB sinC=(bc)/(a^(2))`, then the triangle is

Options

A. Right angled triangle
B. Obtuse angled triangle
C. Equilateral triangle
D. Acute angled triangle

Explanation

To solve the problem, we start with the given equation:

\[ \sin B \sin C = \frac{bc}{a^2} \]

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

### Step 1: Rewrite the equation using the Law of Sines

From the Law of Sines, we know that:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]

This implies:

\[ \sin B = \frac{b \sin A}{a} \quad \text{and} \quad \sin C = \frac{c \sin A}{a} \]

### Step 2: Substitute \(\sin B\) and \(\sin C\) into the equation

Substituting these into the original equation gives:

\[ \left(\frac{b \sin A}{a}\right) \left(\frac{c \sin A}{a}\right) = \frac{bc}{a^2} \]

### Step 3: Simplify the left-hand side

This simplifies to:

\[ \frac{bc \sin^2 A}{a^2} = \frac{bc}{a^2} \]

### Step 4: Cancel \(bc\) from both sides

Assuming \(bc \neq 0\) (which is true since they are sides of a triangle), we can cancel \(bc\):

\[ \sin^2 A = 1 \]

### Step 5: Solve for \(\sin A\)

Taking the square root of both sides, we find:

\[ \sin A = 1 \]

### Step 6: Determine the angle \(A\)

The only angle for which \(\sin A = 1\) is:

\[ A = 90^\circ \]

### Conclusion

Since angle \(A\) is \(90^\circ\), triangle \(ABC\) is a right-angled triangle.

### Final Answer

The triangle is a right-angled triangle.

---