In figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
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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