Có \(\widehat{CMA}+\widehat{CMB}=180^0\) (Hai góc kề bù)
\(\Leftrightarrow5\widehat{CMA}+\widehat{CMA}=180^0\Leftrightarrow\widehat{CMA}=30^0\)
\(\Rightarrow\widehat{BMC}=5.30^0=150^0\)
Có \(\widehat{CMA}+\widehat{AMD}=180^0\)
\(\Leftrightarrow\widehat{AMD}=180^0-30^0=150^0\)
Có \(\widehat{DMB}=\widehat{AMC}=150^0\) (Hai góc đối đỉnh)
Vậy...
Ta có: \(\widehat{BMC}+\widehat{CMA}=180^0\)(hai góc kề bù)
\(\Leftrightarrow5\cdot\widehat{CMA}+\widehat{CMA}=180^0\)
hay \(\widehat{CMA}=30^0\)
\(\Leftrightarrow\widehat{BMC}=5\cdot30^0\)
hay \(\widehat{BMC}=150^0\)
Ta có: \(\widehat{CMA}=30^0\)(cmt)
mà \(\widehat{CMA}=\widehat{BMD}\)(hai góc đối đỉnh)
nên \(\widehat{BMD}=30^0\)
Ta có: \(\widehat{BMC}=150^0\)(cmt)
mà \(\widehat{BMC}=\widehat{AMD}\)(hai góc đối đỉnh)
nên \(\widehat{AMD}=150^0\)
Vậy: \(\widehat{CMA}=30^0\); \(\widehat{BMC}=150^0\); \(\widehat{BMD}=30^0\); \(\widehat{AMD}=150^0\)
