In the given figure, \(PA\) is a tangent from an external point \(P\) to a circle with centre \(O\). If \(\angle POB = 125^{\circ}\), then \(\angle APO\) is equal to :
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Alternatively, find \(\angle POA\) using linear pair: \(\angle POA = 180^{\circ} - 125^{\circ} = 55^{\circ}\).
Then in \(\Delta OAP\), sum of angles is \(180^{\circ}\): \(90^{\circ} + 55^{\circ} + \angle APO = 180^{\circ} \Rightarrow \angle APO = 35^{\circ}\).
To solve this problem, we need to use the concept of tangents and angles related to circles.
Given:
\(\angle POB = 125^{\circ}\)
\(PA\) is a tangent to the circle at point \(A\).
We need to determine \(\angle APO\).
Key concept: The angle between the tangent and the chord through the point of contact is equal to the angle subtended by the chord in the alternate segment.
Since \(PA\) is the tangent and \(AB\) is the chord, \(\angle APO = \angle ABO\).
In \(\triangle OAB\), we know:
\(\angle AOB = 125^{\circ}\) (Central Angle)
The sum of angles in a triangle is \(180^{\circ}\).
Let \(\angle OAB = \angle ABO = x\) (since \(\triangle OAB\) is isosceles, \(OA = OB\)).
Then:
\(x + x + 125^{\circ} = 180^{\circ}\)
Simplifying, we get:
\(2x + 125^{\circ} = 180^{\circ}\)
\(2x = 180^{\circ} - 125^{\circ}\)
\(2x = 55^{\circ}\)
\(x = 27.5^{\circ}\)
Finally, \(\angle APO = \angle ABO = 27.5^{\circ}\).
However, we need to verify where the calculation went wrong:
The mistake was identifying the alternate angle due to a misunderstanding in solving the equation. Re-evaluating gives:
Let’s use the correct angle properties to conclude:
If \(\angle POB = 125^{\circ}\), by properties of tangent and alternate segment, the actual angle \(\angle APO\) calculated around thus calculation is corrected:
The arc calculates \(\angle APO = 180^{\circ} - 125^{\circ}\) is \(\angle ABO\) resulting in \(\angle APO = 55^{\circ}\).
By checking calculation precisely without solving flaws from mistaken assumptions of triangle isometrics we adjust: hence,
