Given a circle with center O and points A, B, and C on the circumference such that angle $\angle AOB$ is $72°$ and angle $\angle BOC$ is also $72°$, what is the measure of angle $\angle CBE$ if E is the point where the extension of $OB$ meets the circle again?

• A. $144°$
• B. $108°$
• C. $36°$
• D. $72°$

Answer: (D)

Verified step by step guidance

1. Identify the given information: Points A, B, C, and E lie on the circle with center O. The central angles \( \angle AOB = 72^\circ \) and \( \angle BOC = 72^\circ \). Point E is on the circle such that \( O, B, E \) are collinear, with E on the opposite side of B from O.
2. Calculate the central angle \( \angle AOC \) by adding \( \angle AOB \) and \( \angle BOC \): \[ \angle AOC = \angle AOB + \angle BOC = 72^\circ + 72^\circ = 144^\circ \]
3. Recall that the measure of an inscribed angle is half the measure of the intercepted arc. Since \( \angle CBE \) is an inscribed angle intercepting the arc \( AC \), its measure is half the measure of the arc \( AC \).
4. Determine the measure of the arc \( AC \). Since \( \angle AOC \) is the central angle intercepting arc \( AC \), the arc \( AC \) measures \( 144^\circ \).
5. Use the inscribed angle theorem to find \( \angle CBE \): \[ \angle CBE = \frac{1}{2} \times \text{arc } AC = \frac{1}{2} \times 144^\circ = 72^\circ \]