Question

In the given figure, PA and PB are tangents to a circle centred at O. If \(\angle AOB = 130^\circ\), then \(\angle APB\) is equal to :

• A. \(130^\circ\)
• B. \(50^\circ\)
• C. \(120^\circ\)
• D. \(90^\circ\)

Hint:

The angle between two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the points of contact at the center.
Simply put:
\[ \angle AOB + \angle APB = 180^\circ \]
So, \(\angle APB = 180^\circ - 130^\circ = 50^\circ\). This mental calculation takes only a few seconds!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given a circle centered at \(O\). The tangents \(PA\) and \(PB\) are drawn from an external point \(P\). The angle subtended by the points of contact at the center, \(\angle AOB\), is \(130^\circ\). We need to determine the value of the angle between the tangents, \(\angle APB\).

Step 2: Key Formula or Approach:
1. The radius of a circle is perpendicular to the tangent at its point of contact. Thus, \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\).
2. The sum of the interior angles of any quadrilateral is \(360^\circ\).
3. In quadrilateral \(OAPB\):
\[ \angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ \]

Step 3: Detailed Explanation:
1. Let \(OAPB\) be the quadrilateral formed by the center \(O\), points of contact \(A\) and \(B\), and the external point \(P\).
2. Identify the perpendicular angles at the points of contact:
\[ \angle OAP = 90^\circ \]
\[ \angle OBP = 90^\circ \]
3. Apply the angle sum property of quadrilaterals:
\[ \angle AOB + \angle OAP + \angle APB + \angle OBP = 360^\circ \]
\[ 130^\circ + 90^\circ + \angle APB + 90^\circ = 360^\circ \]
4. Simplify the equation:
\[ 310^\circ + \angle APB = 360^\circ \]
\[ \angle APB = 360^\circ - 310^\circ = 50^\circ \]
Thus, the angle \(\angle APB\) is \(50^\circ\).

Step 4: Final Answer:
The angle \(\angle APB\) is \(50^\circ\), which corresponds to option (B).