Question -
Answer -
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 parallellines are equal]
Hence, ∠KPG=180 – 60 = 120°
=> ∠GPK = ∠AKL= 120° [Corresponding angles of parallellines are equal]
∠AKH = ∠KHD = 25° [alternate angles of parallellines]
Therefore, ∠HKL = ∠AKH + ∠AKL = 25 + 120 = 145°