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.
210o
B.
150o
C.
105o
D.
50o
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 = 12600, 6 angles = 1110o
Remaining 3 angles = 1260o - 1110o = 150o
size of one of the3 angles \(\frac{150}{3}\) = 50o