Discussion

Explanation:

Given that AB = BC = 2CH = 2CD = EH = EK = 2HK = 4KL = 2LM = MN

And EO = FP

Also,

2CD = EH

EO = FP = CD

∴ KL = PG = $\frac{CD}{2}$

FP = CD : PG = $\frac{CD}{2}$ : ∠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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