If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of `70^(@)`, then `/_POA` is equal to

Options

A. `70^(@)`
B. `55^(@)`
C. `100^(@)`
D. `40^(@)`

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 \( 70^\circ \).

### Step-by-Step Solution:

1. **Draw the Circle and Tangents**:
- Draw a circle with center \( O \).
- Mark a point \( P \) outside the circle.
- Draw two tangents \( PA \) and \( PB \) from point \( P \) to the circle, touching the circle at points \( A \) and \( B \) respectively.

2. **Identify the Given Angle**:
- We are given that the angle between the tangents \( PA \) and \( PB \) is \( \angle APB = 70^\circ \).

3. **Use the Properties of Tangents**:
- Recall that the radius of the circle is perpendicular to the tangent at the point of contact. Therefore, \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

4. **Form a Quadrilateral**:
- Consider the quadrilateral \( OAPB \). The sum of the angles in any quadrilateral is \( 360^\circ \).
- We can express the angles as follows:
\[ \angle OAP + \angle APB + \angle OBP + \angle AOB = 360^\circ \]
- Substituting the known values:
\[ 90^\circ + 70^\circ + 90^\circ + \angle AOB = 360^\circ \]

5. **Solve for \( \angle AOB \)**:
- Combine the known angles:
\[ 90 + 70 + 90 = 250^\circ \]
- Therefore,
\[ 250^\circ + \angle AOB = 360^\circ \]
- Solving for \( \angle AOB \):
\[ \angle AOB = 360^\circ - 250^\circ = 110^\circ \]

6. **Find \( \angle POA \)**:
- The angle \( \angle POA \) is half of \( \angle AOB \) because \( PA \) and \( PB \) are tangents from point \( P \) to the circle.
- Therefore,
\[ \angle POA = \frac{1}{2} \angle AOB = \frac{110^\circ}{2} = 55^\circ \]

### Final Answer:
Thus, \( \angle POA = 55^\circ \).