Given,
AB ‖ CD ‖ EF and GH ‖ KL
Produce HG to M and KL to N
∠MHD and ∠CHG = 60°(Vertically opposite angle)
Since,
MG ‖ NL and transversal cuts them
So,
∠MHD + ∠1 = 180°(Interior angles)
60° + ∠1 = 180°
∠1 = 120°
∠3 = ∠HKD = 25°(Alternate angles) (i)
∠1 = ∠MKL = 120°(Corresponding angles) (ii)
Now,
∠HKL = ∠3 + ∠MKL
= 25° + 120°
= 145°