If in triangle `A B C ,sumsinA/2=6/5a n dsumI I_1=9` (where `I_1,I_2a n dI_3` are excenters and `I` is incenter, then circumradius `R` is equal to `(15)/8` (b) `(15)/4` (c) `(15)/2` (d) `4/(12)`

Options

A. `(15)/(8)`
B. `(15)/(4)`
C. `(15)/(2)`
D. `(4)/(12)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We are given:
- \( \sum \sin \frac{A}{2} = \frac{6}{5} \)
- \( \sum I I_1 = 9 \)
Where \( I_1, I_2, I_3 \) are the excenters and \( I \) is the incenter.

### Step 2: Use the relationship between the incenter and excenters
From the properties of the triangle, we know that:
\[ I \cdot I_1 = 4R \sin \frac{A}{2} \]
Where \( R \) is the circumradius.

### Step 3: Apply the sum notation
Taking the sum on both sides, we have:
\[ \sum I I_1 = \sum (4R \sin \frac{A}{2}) \]
This simplifies to:
\[ 9 = 4R \sum \sin \frac{A}{2} \]

### Step 4: Substitute the known values
We substitute \( \sum \sin \frac{A}{2} = \frac{6}{5} \) into the equation:
\[ 9 = 4R \cdot \frac{6}{5} \]

### Step 5: Solve for \( R \)
Rearranging the equation gives:
\[ R = \frac{9 \cdot 5}{4 \cdot 6} \]
Calculating this:
\[ R = \frac{45}{24} = \frac{15}{8} \]

### Conclusion
Thus, the circumradius \( R \) is equal to \( \frac{15}{8} \).

### Final Answer
The correct option is (a) \( \frac{15}{8} \).

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