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Here, AB ∥ CD and EH is transversal.

∠EFB + ∠BFG = 180° (Linear pair)

∠BFG = 180°- 133°
∠BFG = 47°

∠BFG =∠DGH (Corresponding Angles)

∠DGH = 47°

Example-2:

In the figure, DE || QR and AP and BP are bisectors of ∠EAB and ∠RBA, respectively.
Find ∠APB.

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