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: Understanding the Question:
The problem asks us to find the angle between two tangents drawn from an external point to a circle. We are given the angle subtended by the points of contact at the center of the circle. The points O (center), A (point of contact), P (external point), and B (point of contact) form a quadrilateral OAPB. The information about a third tangent is extra and not needed to find \(\angle APB\).
Step 2: Key Formula or Approach:
We will use two fundamental properties of circles and tangents:
1. The tangent at any point of a circle is perpendicular to the radius through the point of contact. This means \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\).
2. The sum of the interior angles of a quadrilateral is \(360^\circ\).
Step 3: Detailed Explanation:
Consider the quadrilateral OAPB formed by the center O, the external point P, and the points of contact A and B.
The sum of the angles in this quadrilateral is \(360^\circ\).
\[ \angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ \] From the property that the radius is perpendicular to the tangent at the point of contact, we have:
\[ \angle OAP = 90^\circ \] \[ \angle PBO = 90^\circ \] We are given that:
\[ \angle AOB = 50^\circ \] Now, we can substitute these values into the sum of angles equation:
\[ 90^\circ + \angle APB + 90^\circ + 50^\circ = 360^\circ \] \[ \angle APB + 230^\circ = 360^\circ \] Solving for \(\angle APB\):
\[ \angle APB = 360^\circ - 230^\circ \] \[ \angle APB = 130^\circ \] Step 4: Final Answer:
The angle \(\angle APB\) is \(130^\circ\).