AB = BC = CD = DE = EF = FG = GA. Then, ∠DAE...
AB = BC = CD = DE = EF = FG = GA. Then, ∠DAE is approximately
Answer & Explanation
∠EAD=α, then, ∠AFG=α and also ∠ACB=α.
Hence ∠CBD=2α (exterior angle to ΔABC)
CB = CD, hence ∠CDB=2α∠FGC=2α (exterior angle to ΔAFG).
Since GF = EF, ∠FEG=2α
Now, ∠DCE=∠DEC=β (say)
Therefore, on ΔDCB,
180∘−(α+β)+2α+2α=180∘ or β=3α
Further ∠EFD=∠EDF=γ (say)
If CD and EF meet at P,
Now in ΔPED, 180∘−5α+γ+2α=180∘ or γ=3α
Therefore, in ΔEFD,
α+2γ=180∘ or α+6α=180∘ or α=26∘
or approximately 25∘