Question
In figure, find $x.$ Further find $\angle\text{BOC},\angle\text{COD}$ and $\angle\text{AOD}.$
✓
Answer
Since $\angle\text{AOD}$ and $\angle\text{BOD}$ form a line pair,
$\angle\text{AOD}+\angle\text{BOD}=180^\circ$
$\angle\text{AOD}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
Given that, $\angle\text{AOD}=(\text{x}+10)^\circ,\angle\text{COD}=\text{x}^\circ,\angle\text{BOC}=(\text{x}+20)^\circ$
$(\text{x}+10)+\text{x}+(\text{x}+20)=180$
$3\text{x}+30=180$
$3\text{x}=180-30$
$3\text{x}=\frac{150}{3}$
$\text{x}=50$ Therefore, $\angle\text{AOD}=(\text{x}+10)$
$=50+10=60$
$\angle\text{COD}=\text{x}=50^\circ$
$\angle\text{COD}=(\text{x}+20)$
$=50+20=70$
$\angle\text{AOD}=60^\circ,\angle\text{COD}=50^\circ,\angle\text{BOC}=70^\circ$
$\angle\text{AOD}+\angle\text{BOD}=180^\circ$
$\angle\text{AOD}+\angle\text{BOC}+\angle\text{COD}=180^\circ$
Given that, $\angle\text{AOD}=(\text{x}+10)^\circ,\angle\text{COD}=\text{x}^\circ,\angle\text{BOC}=(\text{x}+20)^\circ$
$(\text{x}+10)+\text{x}+(\text{x}+20)=180$
$3\text{x}+30=180$
$3\text{x}=180-30$
$3\text{x}=\frac{150}{3}$
$\text{x}=50$ Therefore, $\angle\text{AOD}=(\text{x}+10)$
$=50+10=60$
$\angle\text{COD}=\text{x}=50^\circ$
$\angle\text{COD}=(\text{x}+20)$
$=50+20=70$
$\angle\text{AOD}=60^\circ,\angle\text{COD}=50^\circ,\angle\text{BOC}=70^\circ$
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