Question

consider a regular hexagon inscribed in circle
hexagons have six congruent sides and six con
figure, angle cab measures 60°.
what is ( mangle acb? )

Explanation:

Step1: Recall the property of a regular hexagon inscribed in a circle

A regular hexagon inscribed in a circle has the property that the side length of the hexagon is equal to the radius of the circle. Let the center of the circle be \(O\). Then, triangles like \(\triangle OAB\) and \(\triangle OBC\) are equilateral triangles.

Step2: Use the inscribed - angle theorem

The inscribed - angle theorem states that the measure of an inscribed angle \(\angle ACB\) is half of the measure of the central angle \(\angle AOB\) that subtends the same arc \(AB\).
Since \(\angle CAB = 60^{\circ}\) and in a regular hexagon, the central angle \(\angle AOB=60^{\circ}\) (because the sum of central angles in a circle is \(360^{\circ}\) and for a regular hexagon, each central angle \(\theta=\frac{360^{\circ}}{6} = 60^{\circ}\)).
We know that in \(\triangle ABC\), \(\angle CAB = 60^{\circ}\), and using the property of angles in a triangle and the fact that \(AB = BC\) (sides of a regular hexagon), \(\triangle ABC\) is an isosceles triangle.
Another approach: The inscribed - angle theorem: If an inscribed angle \(\angle ACB\) subtends an arc \(AB\) and the central angle \(\angle AOB\) subtends the same arc \(AB\), then \(\angle ACB=\frac{1}{2}\angle AOB\). But also, since \(AB = BC\) (sides of a regular hexagon), in \(\triangle ABC\), \(\angle CAB=\angle CBA = 60^{\circ}\).
Using the angle - sum property of a triangle (\(\angle CAB+\angle CBA+\angle ACB = 180^{\circ}\)), substituting \(\angle CAB = 60^{\circ}\) and \(\angle CBA=60^{\circ}\)

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Answer:

\(60\)