Given: y:z = 3:7
Let, y = 3k, therefore z = 7k
y = ∠1 = 3k [∵ Vertically Opposite Angles]
Given: CD||EF,
Therefore,
∠1 + z = 180° [∵ Sum of co-interior angles]
⇒ 3k + 7k = 180°
⇒ 10k = 180°
⇒ k = 180°/10 = 18°
Hence, z = 7k = 7 × 18° = 126°
Given that : AB||CD and CD||EF, therefore AB||EF
⇒ x = z = 126° [∵ Alternate Angles]

Contruction: Produce PQ, so that it intersect ST at M.
Given that: PQ||ST, therefore
∠1 = ∠2 [∵ Correspomding Angles]
⇒ ∠2 = 130°
∠2 + ∠3 = 180° [∵ Linear pair]
⇒ 130° + ∠3 = 180°
⇒ ∠3 = 180° – 130° = 50°
∠5 + ∠4 = 180° [∵ Linear pair]
⇒ 110° + ∠4 = 180°
⇒ ∠4 = 180° – 110° = 70°
In triangle QMR,
∠3 + ∠4 + ∠R = 180°
⇒ 50° + 70° + ∠R = 180°
⇒ 120° + ∠R = 180°
⇒ ∠R = 180° – 120° = 60°