Question:
In figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
Solution:
We have,
∠ACB = 40°; ∠DPB = 120°
∴ ∠APB = ∠DCB = 40° [Angle in same segment]
In ΔPOB, by angle sum property
∠PDB + ∠PBD + ∠BPD = 180°
⇒ 40° + ∠PBD + 120° = 180°
⇒ ∠PBD = 180° − 40° − 120°
⇒ ∠PBD = 20°
∴ ∠CBD = 20°
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