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Question
In the given figure, if ∠ACB = 40°, ∠DPB = 120°, find ∠CBD.
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Solution
It is given that ∠ACB = 40° and ∠DPB = 120°
Construction: Join the point A and B
`angle ACB = angleADB = 40°` (Angle in the same segment)
Now in \[\bigtriangleup BDP\] we have
\[\angle DPB + \angle PBD + \angle BDP = 180° \]
\[ \Rightarrow 120° + \angle PBD + 40° = 180° \]
\[ \Rightarrow \angle PBD = 20° \]
Hence `angle CBD = 20° `
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