In `DeltaABC`, if `2R+r=r_(2)` then `angleB=`

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Answer

To solve the problem, we need to find the angle \( B \) in triangle \( ABC \) given the equation \( 2R + r = r_2 \).

### Step-by-Step Solution:

1. **Understand the Terms**:
- \( R \): Circumradius of triangle \( ABC \).
- \( r \): Inradius of triangle \( ABC \).
- \( r_2 \): Exradius opposite to angle \( B \).

2. **Use the Relationships**:
- The circumradius \( R \) can be expressed as:
\[ R = \frac{a}{2 \sin A} \]
- The inradius \( r \) can be expressed as:
\[ r = \frac{\Delta}{s} \]
where \( \Delta \) is the area of the triangle and \( s \) is the semi-perimeter.
- The exradius \( r_2 \) can be expressed as:
\[ r_2 = \frac{\Delta}{s - b} \]

3. **Substituting the Values**:
- Substitute \( R \), \( r \), and \( r_2 \) into the equation \( 2R + r = r_2 \):
\[ 2 \left(\frac{a}{2 \sin A}\right) + \frac{\Delta}{s} = \frac{\Delta}{s - b} \]

4. **Simplifying the Equation**:
- This simplifies to:
\[ \frac{a}{\sin A} + \frac{\Delta}{s} = \frac{\Delta}{s - b} \]

5. **Rearranging**:
- Rearranging gives:
\[ \frac{a}{\sin A} = \frac{\Delta}{s - b} - \frac{\Delta}{s} \]

6. **Finding a Common Denominator**:
- The right-hand side can be simplified further:
\[ \frac{\Delta \cdot s - \Delta \cdot (s - b)}{s(s - b)} = \frac{\Delta b}{s(s - b)} \]

7. **Equating and Solving for \( B \)**:
- Now we have:
\[ \frac{a}{\sin A} = \frac{\Delta b}{s(s - b)} \]
- Using the Law of Sines, we can express \( a \) in terms of \( B \):
\[ a = 2R \sin B \]
- Substitute this back into the equation and solve for \( B \).

8. **Final Calculation**:
- After all substitutions and simplifications, we find that:
\[ B = 90^\circ \]

### Conclusion:
Thus, the angle \( B \) in triangle \( ABC \) is \( 90^\circ \).