Let O be the in-centre of a triangle ABC and D be a point on the side BC of `Delta ABC`, such that `OD_|_BC` If `angle BOD=15^(@)`, then `angle ABC=?`

Answer

To solve the problem, we need to determine the angle ABC given the conditions provided. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Identify the Incenter and Given Angles**:
- Let O be the incenter of triangle ABC.
- We know that OD is perpendicular to BC, which means angle ODB = 90°.
- We are given that angle BOD = 15°.

2. **Calculate Angle AOD**:
- In triangle BOD, we can find angle AOD using the fact that the sum of angles in a triangle is 180°.
- We have:
\[ \text{Angle BOD} + \text{Angle ODB} + \text{Angle AOD} = 180° \]
- Substituting the known values:
\[ 15° + 90° + \text{Angle AOD} = 180° \]
- Simplifying this gives:
\[ \text{Angle AOD} = 180° - 105° = 75° \]

3. **Relate Angle AOD to Angle ABC**:
- Since O is the incenter, angle AOD is equal to half of angle ACB (because the angle bisector divides the angle into two equal parts).
- Therefore, we have:
\[ \text{Angle ACB} = 2 \times \text{Angle AOD} = 2 \times 75° = 150° \]

4. **Use the Triangle Angle Sum Property**:
- In triangle ABC, the sum of the angles is 180°:
\[ \text{Angle A} + \text{Angle B} + \text{Angle C} = 180° \]
- We know angle C (which is angle ACB) is 150°, so:
\[ \text{Angle A} + \text{Angle B} + 150° = 180° \]
- This simplifies to:
\[ \text{Angle A} + \text{Angle B} = 30° \]

5. **Determine Angle ABC**:
- Since angle A and angle B are supplementary to angle C, we can express angle B in terms of angle A:
\[ \text{Angle B} = 30° - \text{Angle A} \]
- However, we need to find the specific value of angle ABC. We can assume that angle A is equal to angle B for simplicity in this case (as the problem does not specify otherwise).
- Thus, we can set:
\[ 2 \times \text{Angle B} = 30° \]
- This gives:
\[ \text{Angle B} = 15° \]

### Final Answer:
Thus, the angle ABC is \( 15° \).