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

A.
210^{o} 
B.
150^{o} 
C.
105^{o} 
D.
50^{o}
Correct Answer: Option D
Explanation
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