WAEC - Mathematics (1995 - No. 40)

In ΔABC above, BC is produced to D, /AB/ = /AC/ and ∠BAC = 50^o. Find ∠ACD

50^o

60^o

65^o

100^o

115^o

ACB = ABC [Base angles of isoceles triangle]
ACB + ABC + 50^o = 180^o[sum of angles in a triangle]
ACB + ABC = 180^o - 50^o = 130^o

ACB

130^o/2

= 65^o

ACD + ACB = 180^o[sum of ∠s on a straight line]
ACD + 65^o = 180^o
ACD = 180^o - 65^o = 115^o