x + 51° = 107° (Alternate interior angles)
x = 107° – 51° = 56°

x + 55 = 180^v (Sum of supplementary angles or co-interior angles)

x = 125°

x = y = 125° (Corresponding angles)

z + ∠EAB = y (Exterior angle property)

z = 125° – 90° = 35°

According to question,

∠AQP = ∠BQR = x

∠AQP + ∠BQR + ∠PQR = 180° (Linear Pair)

2x + 108° = 180°

x = 36°

Construction: Extend the line CD such that it intersect AO and is parallel to AB.
x = 60° (Corresponding angles)
x + y + 180° – 110° = 180° (Angle sum property)
y = 110° – 60° = 50°

Extend line CD which intersect AO at M.

∠CMO = ∠BAO = 100⁰   (Corresponding angle)

In △MOC

∠MOC + ∠CMO = ∠DCO (exterior angle is equal to the sum of two opposite interior angles)

∠DCO = 100° + 30° = 130°

OBCA is a quadrilateral

∠OAC + ∠BOA + ∠ACB + ∠CBO = 360°

γ + β + ∠ACB + α = 360°

∠ACB = 360° – γ – β – α

x = 360° – ∠ACB

x = γ + β + α

In △ABC
∠ABC + ∠BAC + ∠ACB = 180° (Angle sum property)
∠ABC = 180° – 90° – 40°
∠ABC = 50°
In △BED
∠BED + ∠EBD + ∠BDE = 180° (Angle sum property)
∠EBD = 180° – 50° – 100°
∠EBD = 30°

∠BCD = ∠ABE = 60 (Vertically opposite angle)
In △EAB
∠EAB + ∠EBA + ∠AEB + 180° (Angle sum property)
50° + 60° + ∠AEB = 180°
∠AEB = 70°

∠ABC = ∠BCD = 45° (Alternate interior angles)

In COD

∠COD + ∠CDO + ∠DCO = 180° (Angle sum property)

∠CDO = 180° – 100° – 45°

∠CDO = 35°

(2x-30) ° = (x+20) ° (corresponding angle)
2x -30 = x +20
2x – x = 30+20
x = 50°