If a chord \(AB\) subtends an angle of \(60^\circ\) at the centre of a circle, then the angle between the tangents at \(A\) and \(B\) is also \(60^\circ\). State True/False and justify.
False.
Step 1: In a circle, when we join the radii \(OA\) and \(OB\) (where \(O\) is the centre), the angle at the centre is given as \(\angle AOB = 60^\circ\).
Step 2: At point \(A\), the tangent is always perpendicular to the radius \(OA\). Similarly, at point \(B\), the tangent is perpendicular to the radius \(OB\).
Step 3: Therefore, the angle formed between the two tangents at points \(A\) and \(B\), say \(\theta\), is related to the angle at the centre by the rule:
\(\theta + \angle AOB = 180^\circ\)
Step 4: Substitute the given value \(\angle AOB = 60^\circ\):
\(\theta + 60^\circ = 180^\circ\)
Step 5: Solve for \(\theta\):
\(\theta = 180^\circ - 60^\circ = 120^\circ\)
Final Answer: The angle between the tangents is \(120^\circ\), not \(60^\circ\). Hence, the given statement is False.