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.