# In Fig. 9.39, AB||DE. Find ∠ACD.

Since,

AB DE

ABC = CDE (Alternate angles)

ABC = 40o

In

A + B + ACB = 180o

30o + 40o + ACB = 180o

ACB = 180o – 70o

= 110o (i)

Now,

ACD + ACB = 180o (Linear pair)

ACD + 110o = 180o [From (i)]

ACD = 180o – 110o

= 70o

Hence, ACD = 70o.

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