Three angles of a nonagon are equal and the sum of six other angles is 1110^o. Calculate the size of one of the equal angles.

210^o

150^o

105^o

50^o

Correct answer is D

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the nonagon = 9

where 3 are equal and 6 other angles = 1110^o

( 2 x 9 - 4)90^o = (18 - 4)90^o

= 14 x 90^o = 1260^o

9 angles = 12600, 6 angles = 1110^o

Remaining 3 angles = 1260^o - 1110^o = 150^o

size of one of the3 angles $\frac{150}{3}$ = 50^o

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Three angles of a nonagon are equal and the sum of six other angles is 1110^o. Calculate the size of one of the equal angles.

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Three angles of a nonagon are equal and the sum of six other angles is 1110o. Calculate the size of one of the equal angles.

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Three angles of a nonagon are equal and the sum of six other angles is 1110^o. Calculate the size of one of the equal angles.