Question: Find angle P.---**Chart/Diagram Description:**
* **Type:** Geometric diagram showing a circle and points/lines related to it.
* **Main Elements:**
* **Circle:** A circle is shown with center labeled 'O'.
* **Points:** Points labeled 'A', 'B', and 'C' are located on the circumference of the circle. Point 'O' is at the center. Point 'P' is located outside the circle.
* **Lines/Segments:**
* Segments OA, OB, OC connect the center O to points A, B, and C on the circle, respectively. These are radii.
* Segments AB, BC, AC are chords of the circle, forming triangle ABC inscribed in the circle.
* Segments PA and PB connect point P to points A and B, respectively. These lines appear to be tangent to the circle at A and B, but this is not explicitly stated.
* **Angles:**
* Angle ∠ACB is shown with a curved arc and labeled with the value "65°".
* Other angles are formed by the various segments, such as ∠OAB, ∠OBA, ∠OAC, ∠OCA, ∠OBC, ∠OCB, ∠AOB, ∠AOC, ∠BOC, ∠APB, ∠PAB, ∠PBA. Only ∠ACB is explicitly labeled with a value.
* **Labels and Annotations:** The points are labeled A, B, C, O, P. The angle ∠ACB is labeled as 65°.
**Textual Information:**
* The labels for points are: A, B, C, O, P.
* An angle is labeled with the value: 65°.
**Note:** The image contains only a geometric diagram and a numerical label for an angle. There is no question stem, options, or other explanatory text present in the image.
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We have a circle with center O and points A, B, C on the circumference. Point P is outside the circle. We're given that angle ACB equals 65 degrees, and we need to find angle P. The red lines PA and PB appear to be tangent lines from point P to the circle.
First, we apply the inscribed angle theorem. The inscribed angle ACB equals 65 degrees and subtends arc AB. According to the theorem, the central angle is twice the inscribed angle. Therefore, the central angle AOB equals 2 times 65 degrees, which is 130 degrees.
Next, we identify that PA and PB are tangent lines from point P to the circle. A key property of tangents is that a radius drawn to the point of tangency is perpendicular to the tangent line. Therefore, angle OAP equals 90 degrees and angle OBP equals 90 degrees.
Now we consider the quadrilateral PAOB. The sum of interior angles in any quadrilateral is 360 degrees. We have angle APB plus angle PAO plus angle AOB plus angle OBP equals 360 degrees. Substituting our known values: angle APB plus 90 degrees plus 130 degrees plus 90 degrees equals 360 degrees. This gives us angle APB plus 310 degrees equals 360 degrees. Therefore, angle APB equals 50 degrees.
We have successfully found that angle P equals 50 degrees. Let me summarize our solution: We started with the given inscribed angle ACB of 65 degrees, used the inscribed angle theorem to find the central angle AOB of 130 degrees, applied the tangent property to establish 90-degree angles at A and B, and finally used the quadrilateral angle sum to calculate angle P as 50 degrees. This problem beautifully demonstrates the relationship between tangent angles and inscribed angles in circle geometry.