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 :
Show Hint
Alternatively, find \(\angle POA\) using linear pair: \(\angle POA = 180^{\circ} - 125^{\circ} = 55^{\circ}\).
Then in \(\Delta OAP\), sum of angles is \(180^{\circ}\): \(90^{\circ} + 55^{\circ} + \angle APO = 180^{\circ} \Rightarrow \angle APO = 35^{\circ}\).
Step 1: Understanding the Concept:
A tangent is perpendicular to the radius at the point of contact. Thus, in \(\Delta OAP\), \(\angle OAP = 90^{\circ}\). Step 2: Detailed Explanation:
From the figure, \(P-O-B\) is a straight line segment.
Thus, \(\angle POA\) and \(\angle AOB\) are related, or specifically, \(\angle POB\) is the exterior angle for \(\Delta OAP\) at vertex \(O\).
Using the property that an exterior angle of a triangle is equal to the sum of its two interior opposite angles:
\[ \angle POB = \angle OAP + \angle APO \]
\[ 125^{\circ} = 90^{\circ} + \angle APO \]
\[ \angle APO = 125^{\circ} - 90^{\circ} = 35^{\circ} \] Step 4: Final Answer:
The measure of \(\angle APO\) is \(35^{\circ}\).
