The magnitude of ∠FGO =
Explanation:
Given that AB = BC = 2CH = 2CD = EH = EK = 2HK = 4KL = 2LM = MN
And EO = FP
Also,
2CD = EH
EO = FP = CD
∴ KL = PG = CD2
FP = CD : PG = CD2 : ∠FPG = 90°
∵ The angle are proportionate to the sides opposite to the angles.
∴ ∠FGO = ∠FGP = tan–1 2
Hence, option (d).
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