Q10 of 100 Page 17

In Fig. 16.139, if ∠ACB=40°, ∠DPB=120°, find ∠CBD.

We have,

∠ACB = 40o


∠DPB = 120o


∠ADB = ∠ACB = 40o (Angle on same segment)


In triangle PDB, by angle sum property


∠PDB + ∠PBD + ∠BPD = 180o


40o + ∠PBD + 120o = 180o


∠PBD = 20o


Therefore,


∠CBD = 20o


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