# In ΔABC, AC=BC and BC is extended upto the point D. If ∠ACD=144°, then let us determine the circular value of each of the angles of ΔABC.

ACD = 144°

DCB is straight line, DCB = 180°

ACB = 180° - 144° = 36°

AC=BC, by opposite angle property

CAB = CBA = x

By Angle sum Property

CAB + CBA + ACB = 180°

2x + 36° = 180°

x = 72°

CAB = CBA = 72° and ACB = 36°

And ACB = 36°

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