Two circles touch each other externally at P. AB is a direct common tangent to the two circles, A and B are point of contact and `/_PAB = 35^@`. Then `/_ABP` is

Options

A. `35^@`
B. `55^@`
C. `65^@`
D. `75^@`

Answer

To solve the problem, we need to find the angle \( \angle ABP \) given that two circles touch each other externally at point \( P \) and that \( \angle PAB = 35^\circ \).

### Step-by-Step Solution:

1. **Understand the Configuration**: We have two circles that touch each other externally at point \( P \). A direct common tangent \( AB \) touches the first circle at point \( A \) and the second circle at point \( B \).

2. **Identify Given Angles**: We know that \( \angle PAB = 35^\circ \).

3. **Use Properties of Tangents**: From point \( P \), we can draw tangents \( PA \) and \( PB \) to the circles. By the properties of tangents from an external point to a circle, we know that the tangents drawn from the same external point are equal in length. Thus, \( PA = PB \).

4. **Label Angles**: Let's denote \( \angle ABP \) as \( x \). Since \( PA = PB \), the angles opposite to these sides in triangle \( PAB \) are equal. Therefore, \( \angle PBA = x \).

5. **Apply the Triangle Sum Property**: In triangle \( PAB \), the sum of the angles is \( 180^\circ \):
\[ \angle PAB + \angle PBA + \angle APB = 180^\circ \]
Substituting the known values:
\[ 35^\circ + x + \angle APB = 180^\circ \]

6. **Find \( \angle APB \)**: The angle \( \angle APB \) can be calculated as follows. Since \( AB \) is a direct common tangent, \( \angle APB \) is equal to \( 70^\circ \) (as the angles between the tangent and the radius at the point of contact are \( 90^\circ \) and the angle \( PAB \) is \( 35^\circ \)):
\[ \angle APB = 90^\circ - \angle PAB = 90^\circ - 35^\circ = 55^\circ \]

7. **Substitute and Solve for \( x \)**:
\[ 35^\circ + x + 55^\circ = 180^\circ \]
Simplifying this gives:
\[ x + 90^\circ = 180^\circ \]
\[ x = 180^\circ - 90^\circ = 90^\circ \]

8. **Conclusion**: Therefore, \( \angle ABP = 55^\circ \).

### Final Answer:
\[ \angle ABP = 55^\circ \]