If R and R' are the circumradii of triangles ABC and OBC, where O is the orthocenter of triangle ABC, then :

Options

A. `R'=(R )/(2)`
B. `R'=2R`
C. `R'=R`
D. `R'=3R`

Answer

To solve the problem, we need to establish the relationship between the circumradii \( R \) and \( R' \) of triangles \( ABC \) and \( OBC \), where \( O \) is the orthocenter of triangle \( ABC \).

### Step-by-Step Solution:

1. **Understand the Circumradius**:
The circumradius \( R \) of triangle \( ABC \) is defined as the radius of the circumcircle that passes through all three vertices \( A \), \( B \), and \( C \). Similarly, \( R' \) is the circumradius of triangle \( OBC \).

2. **Properties of the Orthocenter**:
The orthocenter \( O \) of triangle \( ABC \) is the point where the three altitudes intersect. The angles formed at the orthocenter have specific relationships with the angles of triangle \( ABC \).

3. **Using the Sine Rule**:
For triangle \( ABC \):
\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R \]
For triangle \( OBC \):
\[ \frac{BC}{\sin BOC} = 2R' \]

4. **Finding Angle \( BOC \)**:
The angle \( BOC \) can be expressed in terms of the angles of triangle \( ABC \):
\[ \angle BOC = 180^\circ - \angle A \]
Therefore, using the sine function:
\[ \sin BOC = \sin(180^\circ - A) = \sin A \]

5. **Applying the Sine Rule in Triangle \( OBC \)**:
From the sine rule in triangle \( OBC \):
\[ \frac{BC}{\sin BOC} = 2R' \]
Substituting for \( \sin BOC \):
\[ \frac{BC}{\sin A} = 2R' \]

6. **Relating \( BC \) to \( R \)**:
Since \( BC \) corresponds to side \( a \) in triangle \( ABC \):
\[ a = 2R \sin A \]
Substituting this into the equation gives:
\[ \frac{2R \sin A}{\sin A} = 2R' \]
Simplifying this yields:
\[ 2R = 2R' \]
Thus, we have:
\[ R' = R \]

### Conclusion:
The circumradius \( R' \) of triangle \( OBC \) is equal to the circumradius \( R \) of triangle \( ABC \). Therefore, the answer is:
\[ R' = R \]