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 \( OBC \) respectively, where \( O \) is the orthocenter of triangle \( ABC \).

### Step-by-Step Solution:

1. **Identify the triangles and their circumradii**:
- Let \( R \) be the circumradius of triangle \( ABC \).
- Let \( R' \) be the circumradius of triangle \( OBC \).

2. **Understand the angles in triangle \( ABC \)**:
- In triangle \( ABC \), the angles are \( A \), \( B \), and \( C \).
- The orthocenter \( O \) is the point where the altitudes of triangle \( ABC \) intersect.

3. **Analyze triangle \( OBC \)**:
- The angle \( \angle BOC \) can be expressed in terms of the angles of triangle \( ABC \):
\[ \angle BOC = 180^\circ - \angle A = \pi - A \]
- This is because the sum of angles in triangle \( ABC \) is \( 180^\circ \).

4. **Apply the Sine Rule in triangle \( ABC \)**:
- According to the Sine Rule:
\[ \frac{a}{\sin A} = 2R \]
- Here, \( a \) is the side opposite angle \( A \).

5. **Apply the Sine Rule in triangle \( OBC \)**:
- For triangle \( OBC \):
\[ \frac{a}{\sin(B + C)} = 2R' \]
- Since \( B + C = \pi - A \), we have:
\[ \sin(B + C) = \sin(\pi - A) = \sin A \]
- Therefore, we can rewrite the equation as:
\[ \frac{a}{\sin A} = 2R' \]

6. **Equate the two expressions**:
- From the two Sine Rule applications, we have:
\[ 2R = \frac{a}{\sin A} = 2R' \]

7. **Conclude the relationship between \( R \) and \( R' \)**:
- By equating the two expressions, we find:
\[ R = R' \]

Thus, the circumradius \( R \) of triangle \( ABC \) is equal to the circumradius \( R' \) of triangle \( OBC \).

### Final Answer:
\[ R = R' \]