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 \(\angle AOP = 180^\circ - 125^\circ = 55^\circ\) using linear pair, then use angle sum property in \(\triangle OAP\).
Step 1: Understanding the Concept:
A tangent is perpendicular to the radius at the point of contact. The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Step 2: Key Formula or Approach:
In \(\triangle OAP \), \( \angle OAP = 90^\circ \).
Angle \( \angle POB \) is the exterior angle to \(\triangle OAP \) at vertex \( O \) (assuming \( B-O-A \) is a straight line/diameter). Step 3: Detailed Explanation:
1. \( \angle OAP = 90^\circ \) (Radius \(\perp\) Tangent).
2. \(\angle POB\) is an exterior angle to \(\triangle OAP\).
\[ \angle POB = \angle OAP + \angle APO \]
3. Substitute the known values:
\[ 125^\circ = 90^\circ + \angle APO \]
4. Solve for \( \angle APO \):
\[ \angle APO = 125^\circ - 90^\circ = 35^\circ \] Step 4: Final Answer:
The measure of \(\angle APO\) is 35°.
