Question:
In the below fig, AB ∥ CD ∥ EF and GH ∥ KL Find ∠HKL
Solution:
Produce LK to meet GF at N.
Now, alternative angles are equal
∠CHG = ∠HGN = 60°
∠HGN = ∠KNF = 60° [Corresponding angles]
Hence, ∠KNG = 180 − 60 = 120
⟹ ∠GNK = ∠AKL = 120° [Corresponding angles]
∠AKH = ∠KHD = 25° [alternative angles]
Therefore, ∠HKL = ∠AKH + ∠AKL = 25 + 120 = 145°
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