Question

In the above figure tangents \(PA\) and \(PB\) from a point \(P\) to a circle with centre \(O\) are inclined to each other at an angle of \(60^\circ\), then \(\angle POA =\) _____.

• A. \(60^\circ\)
• B. \(70^\circ\)
• C. \(80^\circ\)
• D. \(30^\circ\)

Hint:

For two tangents drawn from an external point: \[ \angle \text{between tangents} + \angle \text{at center} = 180^\circ \] Also remember: \[ \text{Radius} \perp \text{Tangent} \] These two properties solve most tangent-circle angle problems very quickly.

Answer 1

The Correct Option is A

Concept: When two tangents are drawn from an external point to a circle, the following important properties are used:
• The tangent at any point of a circle is perpendicular to the radius through the point of contact.
• Tangents drawn from the same external point are equal in length.
• The line joining the center of the circle to the external point bisects the angle between the tangents. These properties help us form right triangles and calculate unknown angles easily.

Step 1: Identify the given angle between the tangents. The angle formed between tangents \(PA\) and \(PB\) is: \[ \angle APB = 60^\circ \] Since the line joining the center \(O\) to the external point \(P\) bisects the angle between the tangents, therefore: \[ \angle APO = \angle OPB \] Hence: \[ \angle APO = \frac{60^\circ}{2} \] \[ \angle APO = 30^\circ \]

Step 2: Use the tangent-radius perpendicular property. Radius \(OA\) is drawn to the point of tangency \(A\). A tangent is always perpendicular to the radius at the point of contact. Therefore: \[ OA \perp PA \] Thus: \[ \angle OAP = 90^\circ \]

Step 3: Consider triangle \( \triangle OAP \). Now in triangle \(OAP\):
• \(\angle OAP = 90^\circ\)
• \(\angle APO = 30^\circ\)
• \(\angle POA = ?\) Using the angle sum property of a triangle: \[ \angle OAP + \angle APO + \angle POA = 180^\circ \] Substitute the known values: \[ 90^\circ + 30^\circ + \angle POA = 180^\circ \]

Step 4: Solve for \(\angle POA\). First add the known angles: \[ 120^\circ + \angle POA = 180^\circ \] Subtract \(120^\circ\) from both sides: \[ \angle POA = 180^\circ - 120^\circ \] \[ \angle POA = 60^\circ \]

Step 5: Verification using another circle property. Another important theorem states: \[ \angle APB + \angle AOB = 180^\circ \] Since: \[ \angle APB = 60^\circ \] therefore: \[ \angle AOB = 180^\circ - 60^\circ = 120^\circ \] The line \(OP\) bisects \(\angle AOB\), therefore: \[ \angle POA = \frac{120^\circ}{2} = 60^\circ \] This confirms our answer. Final Answer: \[ \boxed{60^\circ} \]