Given,

AB ‖ CD ‖ EF and GH ‖ KL

Produce HG to M and KL to N

∠MHD and ∠CHG = 60^o (Vertically opposite angle)

Since,

MG ‖ NL and transversal cuts them

So,

∠MHD + ∠1 = 180^o (Interior angles)

60^o + ∠1 = 180^o

∠1 = 120^o

∠3 = ∠HKD = 25^o (Alternate angles) (i)

∠1 = ∠MKL = 120^o (Corresponding angles) (ii)

Now,

∠HKL = ∠3 + ∠MKL

= 25^o + 120^o

= 145^o