Find the size of each lettered angle in the following figure.

As CD || BA [Given]

We can write ED || BA

⇒ ∠ECB = ∠CBA [Alternate angles are equal]

⇒ ∠CBA = 75°

Since, ABCD is a parallelogram, we have

⇒ ∠DAB + ∠CBA = 180° (AD || BC, sum of co-int ∠s = 180°)

⇒ (x + 66°) + 75° = 180°

⇒ x + 141° = 180°

⇒ x = 180° – 141°

⇒ x = 39° … (i)

Now, in ∆AMB

⇒ x + 30° + ∠AMB = 180° [Angles sum property of a triangle]

39° + 30° + ∠AMB = 180° [From (i)]

69° + ∠AMB + 180°

∠AMB = 180° – 69° = 111°

Since, ∠AMB = y [Vertically opposite angles are equal]

⇒ y = 111°

Hence, x = 39° and y = 111°