Question

In the figure, If $\angle ACB={40}^{\circ }.\angle DPB={120}^{\circ },$ find $\angle CBD$.

Answer 1

Are AB subtend $\angle ACBand\angle ADB$ in the same segment of a circle

$\therefore \angle ACB=\angle ADB={40}^{\circ }$

In $\Delta PDB$,

$\angle DPB+\angle PBD+\angle ADB={180}^{\circ }$ (Sum of angles of a triangle)

$\Rightarrow {120}^{\circ }+\angle PBD+{40}^{\circ }={180}^{\circ }$

$\Rightarrow {160}^{\circ }+\angle PBD={180}^{\circ }$

$\Rightarrow \angle PBD={180}^{\circ }-{160}^{\circ }={20}^{\circ }$

$\Rightarrow \angle CBD={20}^{\circ }$