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Extend LK to meet line GF at point P.
From figure, CD || GF, so, alternate angles are equal.
∠CHG =∠HGP = 60°
∠HGP =∠KPF = 60° [Corresponding angles of parallel lines are equal]
Hence, ∠KPG =180 – 60 = 120°
=> ∠GPK = ∠AKL= 120° [Corresponding angles of parallel lines are equal]
∠AKH = ∠KHD = 25° [alternate angles of parallel lines]
Therefore, ∠HKL = ∠AKH + ∠AKL = 25 + 120 = 145°
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