In any triangle `ABC` ,the distance of the orthocentre from the vertices A, B,C are in the ratio

Options

A. sin A : sin B : sin C
B. cos A : cos B : cos C
C. tan A : tan B: tan C
D. none of these

Answer

To solve the problem of finding the ratio of the distances from the orthocenter \( O \) of triangle \( ABC \) to its vertices \( A \), \( B \), and \( C \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Orthocenter**:
The orthocenter \( O \) of a triangle is the point where the three altitudes intersect. It is important to know that the distances from the orthocenter to the vertices are related to the angles of the triangle.

2. **Define Distances**:
Let:
- \( OA \) be the distance from the orthocenter \( O \) to vertex \( A \).
- \( OB \) be the distance from the orthocenter \( O \) to vertex \( B \).
- \( OC \) be the distance from the orthocenter \( O \) to vertex \( C \).

3. **Using the Formula for Distances**:
The distances from the orthocenter to the vertices can be expressed in terms of the circumradius \( R \) and the angles of the triangle:
- \( OA = 2R \cos A \)
- \( OB = 2R \cos B \)
- \( OC = 2R \cos C \)

4. **Setting Up the Ratio**:
We can now set up the ratio of these distances:
\[ OA : OB : OC = 2R \cos A : 2R \cos B : 2R \cos C \]

5. **Simplifying the Ratio**:
Since \( 2R \) is a common factor in all three terms, we can simplify the ratio:
\[ OA : OB : OC = \cos A : \cos B : \cos C \]

6. **Conclusion**:
Therefore, the ratio of the distances from the orthocenter to the vertices \( A \), \( B \), and \( C \) is:
\[ OA : OB : OC = \cos A : \cos B : \cos C \]

### Final Answer:
The distances of the orthocenter from the vertices \( A \), \( B \), and \( C \) are in the ratio \( \cos A : \cos B : \cos C \).

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