If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of `110^@` , find `anglePOA`.

Answer

To solve the problem, we need to find the angle \( \angle POA \) given that the tangents \( PA \) and \( PB \) from point \( P \) to the circle with center \( O \) are inclined to each other at an angle of \( 110^\circ \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a circle with center \( O \).
- Point \( P \) is outside the circle, and tangents \( PA \) and \( PB \) touch the circle at points \( A \) and \( B \) respectively.
- The angle between the tangents \( PA \) and \( PB \) is given as \( 110^\circ \).

2. **Using the Properties of Tangents**:
- The radius of the circle is perpendicular to the tangent at the point of contact. Therefore, \( OA \perp PA \) and \( OB \perp PB \).
- This means that \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

3. **Analyzing the Triangles**:
- We can analyze triangles \( AOP \) and \( BOP \).
- Both triangles share the side \( OP \), and since \( OA = OB \) (radii of the same circle) and \( \angle OAP = \angle OBP = 90^\circ \), we can conclude that triangles \( AOP \) and \( BOP \) are congruent by the RHS (Right angle-Hypotenuse-Side) congruence criterion.

4. **Finding the Angles**:
- Since triangles \( AOP \) and \( BOP \) are congruent, we have \( \angle AOP = \angle BOP \).
- The angle \( \angle APB = 110^\circ \) can be expressed as:
\[ \angle APB = \angle AOP + \angle BOP \]
- Since \( \angle AOP = \angle BOP \), we can let \( \angle AOP = \angle BOP = x \).
- Therefore, we have:
\[ x + x = 110^\circ \implies 2x = 110^\circ \implies x = 55^\circ \]

5. **Finding \( \angle POA \)**:
- In triangle \( AOP \), the angles sum up to \( 180^\circ \):
\[ \angle POA + \angle OAP + \angle AOP = 180^\circ \]
- We know \( \angle OAP = 90^\circ \) and \( \angle AOP = 55^\circ \):
\[ \angle POA + 90^\circ + 55^\circ = 180^\circ \]
- Simplifying this gives:
\[ \angle POA + 145^\circ = 180^\circ \]
- Therefore:
\[ \angle POA = 180^\circ - 145^\circ = 35^\circ \]

### Final Answer:
\[ \angle POA = 35^\circ \]