In the figure, if $\angle ACB = 40^o, \angle DPB = 120^o$, find $\angle CBD$.
"
Given:
$\angle ACB = 40^o, \angle DPB = 120^o$.
To do:
We have to find $\angle CBD$.
Solution:
Arc $AB$ subtends $\angle ACB$ and $\angle ADB$ in the same segment of a circle.
Therefore,
$\angle ACB = \angle ADB = 40^o$
In $\triangle PDB$,
$\angle DPB + \angle PBD + \angle ADB = 180^o$ (Sum of angles of a triangle)
$120^o + \angle PBD + 40^o = 180^o$
$160^o + \angle PBD = 180^o$
$\angle PBD = 180^o - 160^o = 20^o$
Hence $\angle CBD = 20^o$.
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