a. Ta có: \(\widehat{MOQ}+\widehat{NOQ}=\widehat{MON}\)
hay \(\widehat{MOQ}+90^o=140^o\) (vì \(OQ\perp ON\) nên \(\widehat{nOQ}=90^o\))
\(\Rightarrow\widehat{MOQ}=50^o\) (1)
và \(\widehat{MOP}+\widehat{NOP}=\widehat{MON}\)
hay \(90^o+\widehat{NOP}=140^o\) (vì \(OP\perp OM\) nên \(\widehat{MOP}=90^o\))
\(\Rightarrow\widehat{NOP}=50^o\) (2)
Từ (1) và (2), suy ra: \(\widehat{MOQ}=\widehat{NOP}\left(=50^o\right)\)
b. Ta có: \(\widehat{POQ}+\widehat{PON}=\widehat{QON}\)
hay \(\widehat{POQ}+50^o=90^o\)
\(\Rightarrow\widehat{POQ}=40^o\)
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