I is the in centre of `Delta ABC ` . If `angleBIC= 108^(@)` then `angleA` =?

Options

A. `54^(@)`
B. `72^(@)`
C. `81^(@)`
D. `36^(@)`

Answer

To solve the problem, we need to find the angle A in triangle ABC given that I is the incenter and angle BIC = 108 degrees.

### Step-by-Step Solution:

1. **Understand the Incenter and Angles**:
The incenter (I) of a triangle is the point where the angle bisectors of the triangle intersect. The angle BIC can be expressed in terms of the angles of triangle ABC. Specifically, the formula is:
\[ \angle BIC = \frac{1}{2} (\angle A + 90^\circ) \]

2. **Substitute the Given Value**:
We know that:
\[ \angle BIC = 108^\circ \]
Therefore, we can set up the equation:
\[ 108^\circ = \frac{1}{2} (\angle A + 90^\circ) \]

3. **Multiply Both Sides by 2**:
To eliminate the fraction, multiply both sides by 2:
\[ 2 \times 108^\circ = \angle A + 90^\circ \]
This simplifies to:
\[ 216^\circ = \angle A + 90^\circ \]

4. **Isolate Angle A**:
Now, we need to isolate angle A by subtracting 90 degrees from both sides:
\[ \angle A = 216^\circ - 90^\circ \]
This gives us:
\[ \angle A = 126^\circ \]

5. **Check for Errors**:
It seems there was an error in the calculation. Let's re-evaluate the equation:
\[ 108^\circ = \frac{1}{2} (\angle A + 90^\circ) \]
Multiplying both sides by 2 gives:
\[ 216^\circ = \angle A + 90^\circ \]
Thus:
\[ \angle A = 216^\circ - 90^\circ = 126^\circ \]

6. **Final Calculation**:
We realize that the angle A cannot be greater than 180 degrees in a triangle. Therefore, we must have made a mistake in the interpretation of the angles. The correct interpretation is:
\[ \angle BIC = 90^\circ + \frac{1}{2} \angle A \]
Thus:
\[ 108^\circ = 90^\circ + \frac{1}{2} \angle A \]
Rearranging gives:
\[ \frac{1}{2} \angle A = 108^\circ - 90^\circ = 18^\circ \]
Therefore:
\[ \angle A = 2 \times 18^\circ = 36^\circ \]

### Conclusion:
The value of angle A is \( \angle A = 36^\circ \).