MCQ
In the given figure, $AB$ and $CD$ are two intersecting chords of a circle. If $\angle\text{CAB}=40^\circ$ and $\angle\text{BCD}=80^\circ,$ then $\angle\text{CBD}=?$
- A$80^\circ $
- B$70^\circ$
- ✓$60^\circ$
- D$50^\circ$
We have:
$\angle\text{CDB}=\angle\text{CAB}=40^\circ$ (Angles in the same segment of a circle)
In $\triangle\text{CBD},$ we have:
$\angle\text{CDB}+\angle\text{BCD}+\angle\text{CBD}=180^\circ$ (Angle sum property of a triangle)
$\Rightarrow40^\circ+80^\circ+\angle\text{CBD}=180^\circ$
$\Rightarrow\angle\text{CBD}=(180^\circ-120^\circ)=60^\circ$
$\Rightarrow\angle\text{CBD}=60^\circ$
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