ABCD is a cyclic quadrilateral. The tangent at A and C meet at a point P. If `angleABC=100^(@)` then `angleAPC` will be euqal to :

Options

A. `10^(@)`
B. `20^(@)`
C. `40^(@)`
D. `80^(@)`

Answer

To solve the problem, we need to use the properties of cyclic quadrilaterals and the angles formed by tangents.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have a cyclic quadrilateral ABCD, and the tangents at points A and C meet at point P. We are given that angle ABC = 100° and need to find angle APC.

2. **Using the Tangent Property**: The tangent to a circle at any point is perpendicular to the radius drawn to the point of tangency. Therefore, the angle between the tangent at point A and the line segment AP (which is a radius to point A) is 90°.

3. **Identifying Angles**: Since angle ABC is given as 100°, we can find angle BAP, which is the angle between line AB and the tangent at A. By the property of tangents, angle BAP = 90° - angle ABC = 90° - 100° = -10°. This indicates that we need to reconsider our approach since angles cannot be negative.

4. **Using Cyclic Quadrilateral Properties**: In a cyclic quadrilateral, opposite angles are supplementary. Therefore, angle ABC + angle ADC = 180°. Since angle ABC = 100°, angle ADC = 180° - 100° = 80°.

5. **Finding Angle APC**: Now, we can use the fact that angle APC is equal to the sum of angles BAP and CAD. Since angle BAP is 90° (from the tangent property) and angle CAD is equal to angle ADC (which is 80°), we have:
\[ \text{angle APC} = \text{angle BAP} + \text{angle CAD} = 90° + 80° = 170°. \]

### Final Answer:
Thus, angle APC = 170°.