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°