Question

PA and PB are tangents to a circle centred at O. If \(\angle PBA = 65^\circ\), then \(\angle APB\) equals :

• A. \(65^\circ\)
• B. \(60^\circ\)
• C. \(50^\circ\)
• D. \(35^\circ\)

Hint:

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 \(180^\circ - 2 \times \angle PBA\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question describes a circle with centre \(O\).
Tangents \(PA\) and \(PB\) are drawn from an external point \(P\) to the circle.
We are given that \(\angle PBA = 65^\circ\) and we need to determine the angle \(\angle APB\).

Step 2: Key Formula or Approach:
We use two important properties of circle tangents:
- Tangents drawn from an external point to a circle are equal in length, so \(PA = PB\).
- In triangle \(\Delta PAB\), since two sides are equal, it is an isosceles triangle, which means the angles opposite to the equal sides are also equal (\(\angle PAB = \angle PBA\)).
- The sum of angles in a triangle is \(180^\circ\).

Step 3: Detailed Explanation:
1. Since \(PA\) and \(PB\) are tangents from the same external point \(P\):
\[ PA = PB \] 2. In triangle \(\Delta PAB\), because \(PA = PB\), the triangle is isosceles.
3. The angles opposite to equal sides must be equal:
\[ \angle PAB = \angle PBA \] 4. We are given \(\angle PBA = 65^\circ\), therefore:
\[ \angle PAB = 65^\circ \] 5. Now, use the angle sum property in \(\Delta PAB\):
\[ \angle APB + \angle PAB + \angle PBA = 180^\circ \] 6. Substitute the known values into the equation:
\[ \angle APB + 65^\circ + 65^\circ = 180^\circ \] \[ \angle APB + 130^\circ = 180^\circ \] 7. Solve for \(\angle APB\):
\[ \angle APB = 180^\circ - 130^\circ = 50^\circ \] 8. Therefore, \(\angle APB\) is \(50^\circ\).

Step 4: Final Answer:
The correct option is (C).