Question
In figure, if $\angle\text{ACB} = 40^\circ, \angle\text{DPB} = 120^\circ,$ find $\angle\text{CBD}.$
✓
Answer
We have,$\angle\text{ACB} = 40^\circ, \angle\text{DPB} = 120^\circ$
$\therefore\angle\text{APB}=\angle\text{DCB}=40^\circ$ [Angle in same segment]
In $\triangle\text{POB},$ by angle sum property$\angle\text{PDB}+\angle\text{PBD}+\angle\text{BPD}=180^\circ$
$\Rightarrow40^\circ +\angle\text{PBD}+120^\circ=180^\circ$
$\Rightarrow\angle\text{PBD}=180^\circ-40^\circ-120^\circ$
$\Rightarrow\angle\text{PBD}=20^\circ$
$\therefore\angle\text{CBD}=20^\circ$
$\therefore\angle\text{APB}=\angle\text{DCB}=40^\circ$ [Angle in same segment]
In $\triangle\text{POB},$ by angle sum property$\angle\text{PDB}+\angle\text{PBD}+\angle\text{BPD}=180^\circ$
$\Rightarrow40^\circ +\angle\text{PBD}+120^\circ=180^\circ$
$\Rightarrow\angle\text{PBD}=180^\circ-40^\circ-120^\circ$
$\Rightarrow\angle\text{PBD}=20^\circ$
$\therefore\angle\text{CBD}=20^\circ$
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