In `aDeltaABC,tanAandtanB` are the roots of `pq(x^(2)+1)=r^(2)x`. Then `Delta` ABC is

Options

A. a right angled triangle
B. an acute angled triangle
C. an obtuse angled triangle
D. an equilateral triangle

Answer

To solve the problem, we need to analyze the given quadratic equation and the properties of triangle ABC.

### Step-by-step Solution:

1. **Identify the Quadratic Equation**:
The given equation is \( pq(x^2 + 1) = r^2 x \). Rearranging this, we get:
\[ pq x^2 - r^2 x + pq = 0 \]

2. **Roots of the Quadratic**:
The roots of this quadratic equation are given as \( \tan A \) and \( \tan B \). According to Vieta's formulas, we know:
- The sum of the roots \( \tan A + \tan B = \frac{r^2}{pq} \)
- The product of the roots \( \tan A \tan B = \frac{pq}{pq} = 1 \)

3. **Using the Product of Roots**:
From the product of the roots, we have:
\[ \tan A \tan B = 1 \]
This implies that:
\[ \tan A = \cot B \quad \text{or} \quad \tan A = \tan(90^\circ - B) \]
Therefore, \( A + B = 90^\circ \).

4. **Finding Angle C**:
Since the angles in a triangle sum up to \( 180^\circ \), we can find angle C:
\[ C = 180^\circ - (A + B) = 180^\circ - 90^\circ = 90^\circ \]

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

### Final Answer:
Triangle ABC is a right-angled triangle.