Đáp án: 

$e) \widehat{yCO}=140^{0}$

$g) \widehat{OAt}=60^{0}$

$h) \widehat{E_{1}}=60^{0}; \widehat{G_{2}}=120^{0}; \widehat{G_{3}}=60^{0}; \widehat{D_{4}}=120^{0}; \widehat{A_{5}}=60^{0}; \widehat{B_{6}}=60^{0}$

Giải thích các bước giải:

`e)` Kẻ $Ob//Ax$
Ta có: $\widehat{xAO}+\widehat{bOA}=180^{0}$ (trong cùng phía)
$\Rightarrow \widehat{bOA}=180^{0}-\widehat{xAO}$
$\Leftrightarrow \widehat{bOA}=180^{0}-50^{0}=130^{0}$
Ta lại có: $\widehat{AOC}+\widehat{AOb}+\widehat{bOC}=360^{0}$
$\Rightarrow \widehat{bOC}=360^{0}-(\widehat{AOC}+\widehat{AOb})$
$\Leftrightarrow \widehat{bOC}=360^{0}-(90^{0}+130^{0})=140^{0}$
Ta lại có: $\widehat{bOC}=\widehat{yCO}(=140^{0})$
`g)` Kẻ $Ob//At//Cx$
Ta có: $\widehat{xCO}+\widehat{bOc}=180^{0}$ (trong cùng phía)
$\Rightarrow \widehat{bOC}=180^{0}-\widehat{xCO}$
$\Leftrightarrow \widehat{bOC}=180^{0}-30^{0}=150^{0}$
Ta lại có: $\widehat{bOC}+\widehat{COA}+\widehat{AOb}=360^{0}$
$\Rightarrow \widehat{AOb}=360^{0}-(\widehat{bOC}+\widehat{COA})$
$\Leftrightarrow \widehat{AOb}=360^{0}-(150^{0}+90^{0})=120^{0}$
Ta lại có: $\widehat{AOb}+\widehat{OAt}=180^{0}$ (trong cùng phía)
$\Rightarrow \widehat{OAt}=180^{0}-\widehat{AOb}$
$\Leftrightarrow \widehat{OAt}=180^{0}-120^{0}=60^{0}$
`h)` Ta có: $\widehat{C_{1}}=\widehat{E_{1}}$ (so le trong)
mà $\widehat{C_{1}}=60^{0}$
$\Rightarrow \widehat{E_{1}}=60^{0}$
Ta lại có: $\widehat{D_{1}}=\widehat{G_{2}}$ (đồng vị)
mà $\widehat{D_{1}}=110^{0}$
$\Rightarrow \widehat{G_{2}}=120^{0}$
Ta lại có: $\widehat{G_{3}}+\widehat{G_{2}}=180^{0}$ (kề bù)
$\Rightarrow \widehat{G_{3}}=180^{0}-\widehat{G_{2}}$
$\Leftrightarrow \widehat{G_{3}}=180^{0}-120^{0}=60^{0}$
Ta lại có: $\widehat{D_{4}}=\widehat{G_{2}}$ (so le trong)
mà $\widehat{G_{2}}=120^{0}$
$\Rightarrow \widehat{D_{4}}=120^{0}$
Ta lại có: $+)\widehat{C_{1}}=\widehat{C_{2}}$ (đối đỉnh)
mà $\widehat{C_{1}}=60^{0}$
$\Rightarrow \widehat{C_{2}}=60^{0}$
           $   +) \widehat{C_{2}}=\widehat{A_{5}}$ (đồng vị)
mà $\widehat{C_{2}}=60^{0}$
$\Rightarrow \widehat{A_{5}}=60^{0}$
Ta lại có: $\widehat{B_{6}}=\widehat{G_{3}}$ (đồng vị)
mà $\widehat{G_{3}}=60^{0}$
$\Rightarrow \widehat{B_{6}}=60^{0}$

e) Kẻ $Oa//Ax$

$⇒Oa//Ax//Cy$

$⇒\widehat{aOA}=\widehat{xAO}=50^o$ (so le trong)

$⇒\widehat{aOC}=90^o-50^o=40^o$

$⇒\widehat{aOC}+\widehat{OCy}=180^o$ (trong cùng phía)

mà $\widehat{aOC}=40^o$

$⇒\widehat{OCy}=180^o-40^o=140^o$

g) Kẻ $Oa//At$

$⇒Oa//At//Cx$

$⇒\widehat{aOC}=\widehat{xCO}=30^o$ (so le trong)

$⇒\widehat{aOA}=90^o-30^o=60^o$

$⇒\widehat{aOA}=\widehat{OAt}=60^o$ (so le trong)

h) Kẻ $Fa//MN$

$⇒Fa//MN//PQ$

$⇒\widehat{aFM}=\widehat{MNF}=45^o$ (so le trong)

$⇒\widehat{aFP}=\widehat{QPF}=53^o$ (so le trong)

$⇒\widehat{MFP}=45^o+53^o=98^o$