To determine if hexagon MNOPQR is equiangular, we start by recalling that each side of the hexagon is congruent, which means all sides are equal in length. This leads to the conclusion that the hexagon is regular, where both sides and angles are congruent.
First, let's figure out the measure of the intercepted arcs created by each side of the hexagon. Since the six congruent sides of the hexagon divide the circle into six equal parts, each arc measures:
[tex]\[ \frac{360^\circ}{6} = 60^\circ \][/tex]
Thus, the measure of each intercepted arc is [tex]\( 60^\circ \)[/tex].
Next, consider that an inscribed angle in a circle is half the measure of its intercepted arc. Since each intercepted arc measures [tex]\( 60^\circ \)[/tex], an inscribed angle that intercepts one of these arcs will measure:
[tex]\[ \frac{60^\circ}{2} = 30^\circ \][/tex]
To show that the hexagon is equiangular, we need to look at the internal angles of the hexagon. Since a regular hexagon is equilateral, all internal angles are congruent. The internal angle at each vertex of the hexagon is an external angle of the triangle formed by two adjacent sides of the hexagon and a radius of the circle.
Each internal angle of the hexagon can be calculated as:
[tex]\[ 180^\circ - 30^\circ = 150^\circ \][/tex]
But since each vertex angle has two equal parts matched by the sides:
[tex]\[ \frac{150^\circ}{2} = 60^\circ \][/tex]
Hence the measure of each internal angle of the hexagon is [tex]\( 60^\circ \)[/tex].
Therefore, since a regular hexagon has equal internal angles, hexagon MNOPQR is confirmed to be equiangular with each internal angle measuring [tex]\( 60^\circ \)[/tex].
