Please help easy stuff...........
2 answers:
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Consider the center of the nonagon. If we join the center to all the vertices, the central angle of 360°, is divided into 9 equal angles, each with a measure of:
360°/9=40°
So to a nonagon rotated 40°, clockwise or counterclockwise, is indistinguishable from the original one.
similarly if we rotate 80 ° c.wise or c.c.wise
in general, a nonagon has a rotational symmetry of 40°n,
where n∈{...-3, -2, -1, 0, 1, 2, 3}...
where 40°n, for n negative, means we are rotating clockwise.
Answer: any angle which is a multiple of 40°
360°/9=40°
So to a nonagon rotated 40°, clockwise or counterclockwise, is indistinguishable from the original one.
similarly if we rotate 80 ° c.wise or c.c.wise
in general, a nonagon has a rotational symmetry of 40°n,
where n∈{...-3, -2, -1, 0, 1, 2, 3}...
where 40°n, for n negative, means we are rotating clockwise.
Answer: any angle which is a multiple of 40°