Question:

\(\angle\)ACB is inscribed angle in a circle with centre O. If \(\angle\)ACB = 65\(^{\circ}\), then what is measure of its intercepted arc AXB ?

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Always remember the key circle theorems: the angle at the center is double the angle at the circumference (inscribed angle), angles in the same segment are equal, and the angle in a semicircle is a right angle (90\(^{\circ}\)).
  • 65\(^{\circ}\)
  • 230\(^{\circ}\)
  • 295\(^{\circ}\)
  • 130\(^{\circ}\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
This question is based on the Inscribed Angle Theorem in circle geometry. The theorem states that the measure of an angle inscribed in a circle is half the measure of its intercepted arc.

Step 2: Key Formula or Approach:
The formula derived from the Inscribed Angle Theorem is:
\[ \text{Measure of Inscribed Angle} = \frac{1}{2} \times \text{Measure of Intercepted Arc} \] Or, rearranging for the arc:
\[ \text{Measure of Intercepted Arc} = 2 \times \text{Measure of Inscribed Angle} \]

Step 3: Detailed Explanation:
We are given the measure of the inscribed angle, \(\angle\)ACB.
\[ \angle\text{ACB} = 65^{\circ} \] The intercepted arc is AXB.
Using the formula from Step 2:
\[ \text{Measure of arc AXB} = 2 \times \angle\text{ACB} \] \[ \text{Measure of arc AXB} = 2 \times 65^{\circ} \] \[ \text{Measure of arc AXB} = 130^{\circ} \]

Step 4: Final Answer:
The measure of the intercepted arc AXB is 130\(^{\circ}\).

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