MCQ
In the given figure, $\angle\text{OAB}=75^\circ, \angle\text{OBA}=55^\circ$ and $\angle\text{OCD}=100^\circ.$ then, $\angle\text{ODC}=?$
- A$20^\circ$
- B$25^\circ$
- ✓$30^\circ$
- D$35^\circ$
✓
Answer
Correct option: C.
$30^\circ$
In $\triangle\text{OAB},$ we have
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ (Angle sum property)
$\Rightarrow55^\circ+75^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=50^\circ$
$\Rightarrow\angle\text{COD}=\angle\text{AOB}=50^\circ$ (Vertivcally opposite angles)
In $\triangle\text{OCD},$ we have
$\angle\text{COD}+\angle\text{OCD}+\angle\text{ODC}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ (Angle sum property)
$\Rightarrow55^\circ+75^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\angle\text{AOB}=50^\circ$
$\Rightarrow\angle\text{COD}=\angle\text{AOB}=50^\circ$ (Vertivcally opposite angles)
In $\triangle\text{OCD},$ we have
$\angle\text{COD}+\angle\text{OCD}+\angle\text{ODC}=180^\circ$ (Angle sum property)
$\Rightarrow50^\circ+100^\circ+\text{x}=180^\circ$
$\Rightarrow\text{x}=30^\circ$
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