In the given figure, if AB, DE are parallel and ∠ABC = 48°, then the value of ∠ACB + 1.5 ∠ACD is?

100° 48° 148° 80°

148°

ATQ, 3a = 48° a = 16°  Now, as per parallel line concept: ⇒ ∠ABC + ∠DCB = 180° ⇒ 3a + 2a + b = 180° ⇒ b = 180° - 80° = 100° = ∠ACB ⇒ Hence, ∠ACB + 1.5∠ACD = b + 1.5(2a) = 100° + 48° = 148°