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

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 = 1110o

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

14 x 90o = 1260o

9 angles = 1260°; 6 angles = 110o

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

Size of one of the 3 angles \(\frac{150}{3}\) = 50o

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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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