# In Fig. 8.127, AB||CD||EF and GH||KL. The measure of ∠HKL isA. 85°B. 135°C. 145°D. 215°

Given,

AB CD EF and GH KL

Produce HG to M and KL to N

MHD and CHG = 60o (Vertically opposite angle)

Since,

MG NL and transversal cuts them

So,

MHD + 1 = 180o (Interior angles)

60o + 1 = 180o

1 = 120o

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

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

Now,

HKL = 3 + MKL

= 25o + 120o

= 145o

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