$$ABCDE$$ is a pentagon inscribed in a circle with centre $$O$$. If $$AB = BC = CD$$ and angle $$ABC$$ is $$132^\circ$$ then angle $$AEB$$ is

Solution

$$\angle ABC = 132^\circ$$ is an inscribed angle on the arc $$AC$$ not containing $$B$$, so that arc is $$264^\circ$$ and the arc $$ABC$$ through $$B$$ is $$360^\circ - 264^\circ = 96^\circ$$. Since $$AB = BC$$, the arcs $$AB$$ and $$BC$$ are $$48^\circ$$ each. $$\angle AEB$$ stands on the arc $$AB$$, so $$\angle AEB = \frac{48^\circ}{2} = 24^\circ$$.

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