The side BC of ΔABC has been produced to a point D, such that ∠ACD = 140°. If ∠B = 3∠A, then find ∠A.
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By exterior angle theorem
AngleACD=AngleA+AngleB
120degree=angleA+1/2angleA
120degree=2angleA+angleB/2
240degree= 3AngleA
angleA=80degree
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