1. In the figure AB=BC=CD=DE=EF=FG=GA, then $$\angle{DAE}$$ is approximately
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By: anil on 05 May 2019 03.05 pm
Let angle EAD be x. So according to given conditions we get , ACB = x. As an external angle CBD = 2x. Also we know that total angle on an line is 180 we get EFD = 3x because of which EDF = 3x. Similarly on opposite side we get AED = 3x. So in total x+3x+3x = 180. We get x = 25 (approx).
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Let angle EAD be x. So according to given conditions we get , ACB = x. As an external angle CBD = 2x. Also we know that total angle on an line is 180 we get EFD = 3x because of which EDF = 3x. Similarly on opposite side we get AED = 3x. So in total x+3x+3x = 180. We get x = 25 (approx).