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Question 34
In a triangle ABC, the side BC is extended up to D. Such that CD = AC, if angleBAD = $$109°$$ and angleACB=$$72°$$ then the value of angleABC is
Solution
Angle ACB =72$$^o$$
Angle ACD = 108$$^o$$
Angle CAD = Angle ADC = 36$$^o$$
Angle BAD = Angle BAC + Angle CAD =109$$^o$$
Angle BAC = 73$$^o$$
Angle ABC = 180 - Angle BAC - Angle ACB = 180 - 73 - 72 = 35$$^o$$
Hence Option A is the correct answer.
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