The distance of the incentre of the triangle ABC from A is

Options

A. `4Rsin(A//2)`
B. `4Rsin](B+C)//2]`
C. `4Rsin(B//2)sin(C//2)`
D. none of these

Answer

To find the distance of the incenter (I) of triangle ABC from vertex A, we can follow these steps:

### Step 1: Understand the Triangle and the Incenter
We have triangle ABC with angles A, B, and C. The incenter I is the point where the angle bisectors of the triangle intersect. The distance we need to find is \(AI\).

### Step 2: Use the Angle Bisector Theorem
In triangle AIC, we know that:
- Angle AIC = \(180^\circ - \frac{A}{2} - \frac{C}{2}\)

### Step 3: Apply the Sine Rule
Using the sine rule in triangle AIC, we have:
\[ \frac{AI}{\sin C/2} = \frac{AC}{\sin AIC} \]
Where:
- \(AC = b\) (opposite angle B)
- \(AI = d\) (the distance we want to find)

This gives us:
\[ AI = \frac{b \cdot \sin C/2}{\sin AIC} \]

### Step 4: Calculate Angle AIC
Substituting the expression for angle AIC:
\[ AIC = 180^\circ - \frac{A}{2} - \frac{C}{2} = \frac{B}{2} \]
Thus, we can rewrite the sine of angle AIC:
\[ \sin AIC = \sin \left( \frac{B}{2} \right) \]

### Step 5: Substitute Back into the Sine Rule
Now substituting back into our sine rule expression:
\[ AI = \frac{b \cdot \sin C/2}{\sin \left( \frac{B}{2} \right)} \]

### Step 6: Final Expression
Using the relationship between the sides and angles of the triangle, we can express \(AI\) in terms of the circumradius \(R\) and the sine of the angles:
\[ AI = 2R \cdot \frac{\sin B}{\sin \left( \frac{B}{2} \right)} \]

### Step 7: Simplify the Expression
Using the double angle identity:
\[ \sin B = 2 \sin \left( \frac{B}{2} \right) \cos \left( \frac{B}{2} \right) \]
We can substitute this back into our expression for \(AI\):
\[ AI = 4R \cdot \sin \left( \frac{B}{2} \right) \cdot \cos \left( \frac{B}{2} \right) \]

### Conclusion
Thus, the distance from the incenter I to vertex A is given by:
\[ AI = 4R \cdot \sin \left( \frac{B}{2} \right) \cdot \sin \left( \frac{C}{2} \right) \]