Question

From an external point P, two tangents PA and PB are drawn to a circle with centre O. Radii OA and OB are perpendicular to the tangents and the angle \(\angle AOB = 50^\circ\). A third tangent is drawn which intersects both PA and PB. Find the angle \(\angle APB\).

Hint:

In the quadrilateral formed by two tangents from an external point and the radii to the points of contact, the angle at the center (\(\angle AOB\)) and the angle between the tangents (\(\angle APB\)) are always supplementary. They add up to \(180^\circ\). So, a quick calculation is \(\angle APB = 180^\circ - \angle AOB\).

Answer 1

Correct Answer: 130

Step 1: Since \(PA=PB\) (tangents from an external point are equal) and \(OA=OB\) (radii), OAPB is a kite, so the diagonal OP bisects both \(\angle AOB\) and \(\angle APB\).
Step 2: This bisection gives \(\angle AOP = \angle BOP = 25^\circ\) (half of \(50^\circ\)). In right triangle OAP (right-angled at A), \(\angle APO = 90^\circ - 25^\circ = 65^\circ\); by symmetry \(\angle BPO = 65^\circ\) too.
Step 3: Add the two halves of \(\angle APB\):
\[ 65^\circ + 65^\circ = \boxed{130^\circ} \]
Final Answer: \(\angle APB = 130^\circ\).