PA and PB are tangents to a circle centred at O. If \(\angle PBA = 65^\circ\), then \(\angle APB\) equals :
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For any pair of tangents \(PA\) and \(PB\), the triangle \(\Delta PAB\) is always isosceles with \(PA = PB\).
Thus, the vertex angle \(\angle APB\) can always be directly calculated as \(180^\circ - 2 \times \angle PBA\).
Step 1: Use the radius-tangent right angles at both points of contact.
Since $PA$ and $PB$ are tangents to the circle at $A$ and $B$, the radii to these points are perpendicular to the tangents:
\[ \angle OAP = 90^{\circ}, \qquad \angle OBP = 90^{\circ} \]
Step 2: Find $\angle OBA$ from the given angle.
We are given $\angle PBA = 65^{\circ}$. Since $\angle OBP = 90^{\circ}$ is made up of $\angle OBA$ and $\angle PBA$ together:
\[ \angle OBA = \angle OBP - \angle PBA = 90^{\circ} - 65^{\circ} = 25^{\circ} \]
Step 3: Use the isosceles triangle $OAB$ to get $\angle AOB$.
Since $OA$ and $OB$ are both radii, $OA = OB$, so triangle $OAB$ is isosceles and $\angle OAB = \angle OBA = 25^{\circ}$.
By the angle sum property:
\[ \angle AOB = 180^{\circ} - 25^{\circ} - 25^{\circ} = 130^{\circ} \]
Step 4: Use the quadrilateral $OAPB$ to find $\angle APB$.
The four angles of quadrilateral $OAPB$ add up to $360^{\circ}$:
\[ \angle OAP + \angle APB + \angle PBO + \angle BOA = 360^{\circ} \]
\[ 90^{\circ} + \angle APB + 90^{\circ} + 130^{\circ} = 360^{\circ} \]
\[ \angle APB = 360^{\circ} - 310^{\circ} = 50^{\circ} \]
