
Pls help to solve (b)
Pls help to solve (b)
(a) Angle AEB = 180-80-20 = 80 => AE = AD = EB
Angle BAC = 60 => Angle AEB = 60 (equilateral triangle AE = EB = AB)
Hence Angle BEC = 180 – 80-60 = 400 (Angles on a straight line)
(b) Angle AEC = 60 + 40 = 100 => Angle ACE = 40 => AE = EC = EB
Angle ECB = (180-40)/ 2 = 70 (angles in isoc triangle, EB=EC)
Therefore, Angle ACB = Angle ECB – Angle ACE = 70-40 = 300
PS: Aaron, please click on “Accept Answer” if the answer is correct.