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.
210o
150o
105o
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
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