In `Delta ABC`, right angled at A, `cos^(-1)((R )/(r_(2)+r_(3)))` is

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `45^(@)`

Explanation

To solve the problem, we need to find the value of \( \cos^{-1}\left(\frac{R}{r_2 + r_3}\right) \) in triangle \( \Delta ABC \) which is right-angled at \( A \).

### Step-by-Step Solution:

1. **Identify the Triangle Properties**:
In triangle \( ABC \), since it is right-angled at \( A \), we know that:
- \( A = 90^\circ \)
- The circumradius \( R \) and the inradii \( r_1, r_2, r_3 \) are related to the angles and sides of the triangle.

2. **Use the Relationship Between Inradii**:
We have the property that:
\[ r_2 + r_3 = 4R \cos^2\left(\frac{A}{2}\right) \]
Since \( A = 90^\circ \), we find \( \frac{A}{2} = 45^\circ \).

3. **Calculate \( \cos^2(45^\circ) \)**:
We know that:
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \quad \Rightarrow \quad \cos^2(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]

4. **Substitute into the Equation**:
Now substituting \( \cos^2(45^\circ) \) into the equation for \( r_2 + r_3 \):
\[ r_2 + r_3 = 4R \cdot \frac{1}{2} = 2R \]

5. **Substitute Back into the Cosine Inverse**:
Now we can substitute \( r_2 + r_3 \) back into our original equation:
\[ \cos^{-1}\left(\frac{R}{r_2 + r_3}\right) = \cos^{-1}\left(\frac{R}{2R}\right) = \cos^{-1}\left(\frac{1}{2}\right) \]

6. **Find the Angle**:
We know that:
\[ \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ \]

### Final Answer:
Thus, the value of \( \cos^{-1}\left(\frac{R}{r_2 + r_3}\right) \) is:
\[ \boxed{60^\circ} \]