If the sines of the angle A and B of a triangle ABC satisfy the equation `c^(2) x^(2) - c (a + b) x + ab = 0`, then the triangle

Options

A. is acute angled
B. is right angled
C. is obtus angled
D. satisfies the equation `sin A + cos A = ((a +b))/(c)`

Explanation

To solve the problem, we need to analyze the given quadratic equation in terms of the sines of angles A and B in triangle ABC. The equation is:

\[ c^2 x^2 - c(a + b)x + ab = 0 \]

### Step-by-Step Solution:

1. **Identify the Roots of the Quadratic Equation:**
The roots of the quadratic equation can be identified using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = c^2 \), \( b = -(c(a + b)) \), and \( c = ab \).

2. **Sum and Product of Roots:**
From Vieta's formulas, the sum of the roots \( x_1 + x_2 \) is given by:
\[ x_1 + x_2 = \frac{c(a + b)}{c^2} = \frac{a + b}{c} \]
The product of the roots \( x_1 \cdot x_2 \) is:
\[ x_1 \cdot x_2 = \frac{ab}{c^2} \]

3. **Relate Roots to Sine Values:**
Since the roots correspond to \( \sin A \) and \( \sin B \), we can write:
\[ \sin A + \sin B = \frac{a + b}{c} \]
\[ \sin A \cdot \sin B = \frac{ab}{c^2} \]

4. **Using the Sine Rule:**
According to the sine rule, we have:
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
Therefore, we can express \( \sin A \) and \( \sin B \) in terms of the sides and circumradius \( R \):
\[ \sin A = \frac{a}{2R}, \quad \sin B = \frac{b}{2R} \]

5. **Substituting into the Equations:**
Substitute these expressions into the sum and product:
\[ \frac{a}{2R} + \frac{b}{2R} = \frac{a + b}{c} \]
This simplifies to:
\[ \frac{a + b}{2R} = \frac{a + b}{c} \]
From this, we can conclude:
\[ c = 2R \]

6. **Finding the Relationship Between Angles:**
Now, substituting into the product of the roots:
\[ \left(\frac{a}{2R}\right) \left(\frac{b}{2R}\right) = \frac{ab}{c^2} \]
This simplifies to:
\[ \frac{ab}{4R^2} = \frac{ab}{(2R)^2} \]
This confirms our previous finding.

7. **Conclusion About Angle C:**
Since \( c = 2R \), we can use the sine rule:
\[ c = 2R \sin C \]
Thus, we have:
\[ 2R = 2R \sin C \implies \sin C = 1 \implies C = 90^\circ \]

### Final Answer:
Therefore, triangle ABC is a right-angled triangle with angle C being \( 90^\circ \).