In the given figure, if AC, DE are parallel and ∠CAB = 38° , then the value of ∠ABC + 5∠CBD is:

158° 178° 218° 196°

218°

We have, ∠CAB = 38 ∠ABC = b ∠CBD = a ∠CAB = 2a According to the question, 2a = 38o ∠CBD = a = 19o ∠CAB + ∠ABD = 180 (Because AC and BD are parallel)  ⇒ 2a + b + a = 180o ⇒ 38 + b + 19 = 180o ⇒ b = 180 - 57 ⇒ b = 123o ∴∠ABC = 123o Now, ∠ABC + 5 ∠CBD = 123 + 5 × 19 = 123 + 95 = 218o