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Question

In the figure, if BAC=60o and BCA=20o, find ADC


Solution

In ΔABC, BAC+ABC+ACB=180o  (Sum of angles of triangle)

60o+ABC+20o=180o

ABC+80o=180o

ABC=180o80o=100o

  ABCD is a cyclic quadrilateral,

ABC+ADC=180o

100o+ADC=180o

ADC=180o100o=80o

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