JAMB - Mathematics (1987 - No. 27)
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
Selitys
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
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