In triangle ABC, if `r_(1)+r_(2)=3R` and `r_(2)+r_(3)=2R`, then

Options

A. `angle A =90^(@)`
B. `angle B=45^(@)`
C. `angle C=60^(@)`
D. triangle ABC is right angled isosceles

Explanation

To solve the problem, we need to analyze the given equations involving the inradii \( r_1, r_2, \) and \( r_3 \) of triangle \( ABC \) and relate them to the circumradius \( R \).

### Step-by-Step Solution:

1. **Understanding the Inradii and Circumradius**:
- The inradii \( r_1, r_2, r_3 \) are given by:
\[ r_1 = \frac{A}{s - a}, \quad r_2 = \frac{A}{s - b}, \quad r_3 = \frac{A}{s - c} \]
where \( A \) is the area of triangle \( ABC \) and \( s \) is the semi-perimeter \( s = \frac{a + b + c}{2} \).

2. **Setting Up the Equations**:
- From the problem, we have:
\[ r_1 + r_2 = 3R \quad (1) \]
\[ r_2 + r_3 = 2R \quad (2) \]

3. **Substituting the Inradii**:
- Substitute the expressions for \( r_1 \) and \( r_2 \) into equation (1):
\[ \frac{A}{s - a} + \frac{A}{s - b} = 3R \]
- Multiply through by \( (s - a)(s - b) \):
\[ A(s - b) + A(s - a) = 3R(s - a)(s - b) \]
- Simplifying gives:
\[ A(2s - a - b) = 3R(s - a)(s - b) \quad (3) \]

4. **Substituting for \( r_2 \) and \( r_3 \)**:
- Substitute \( r_2 \) and \( r_3 \) into equation (2):
\[ \frac{A}{s - b} + \frac{A}{s - c} = 2R \]
- Multiply through by \( (s - b)(s - c) \):
\[ A(s - c) + A(s - b) = 2R(s - b)(s - c) \]
- Simplifying gives:
\[ A(2s - b - c) = 2R(s - b)(s - c) \quad (4) \]

5. **Relating Areas and Circumradius**:
- We know from the area formula:
\[ A = \frac{abc}{4R} \]
- Substitute this into equations (3) and (4) to relate \( R \) and the sides of the triangle.

6. **Finding Angles**:
- From the relationships established, we can use the cosine rules:
\[ \cos \frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}, \quad \cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}}, \quad \cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}} \]
- Using the derived relationships, we can find \( \cos \frac{C}{2} = \frac{\sqrt{3}}{2} \), which implies \( C = 60^\circ \).

7. **Finding Remaining Angles**:
- If \( C = 60^\circ \) and using the sum of angles in a triangle, we can find \( A \) and \( B \):
\[ A + B + C = 180^\circ \]
- Solving gives \( A = 90^\circ \) and \( B = 30^\circ \).

### Final Result:
- The angles of triangle \( ABC \) are:
\[ A = 90^\circ, \quad B = 30^\circ, \quad C = 60^\circ \]