Ex 6.3, 3 - In figure, if AB || DE, ∠BAC = 35° & ∠CDE = 53° - Triangle - Problems

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Ex6.3, 3 In the given figure, if AB || DE, BAC = 35 and CDE = 53 , find DCE. Since AB || DE and AE is a transversal. BAC = CED 35 = CED CED = 35 In CDE, CDE + CED + DCE = 180 53 + 35 + DCE = 180 88 + DCE = 180 DCE = 180 88 DCE = 92

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