Xét ΔABC, có: $\widehat{BAC}$ + $\widehat{ABC}$ + $\widehat{ACB}$ = $180^o$
⇒ $\widehat{ACB}$ = $180^o$ - $(\widehat{BAC}$ + $\widehat{ABC})$
⇒ $\widehat{ACB}$ = $180^o$ - $(50^o$ + $70^o)$
= $180^o$ - $120^o$
= $60^o$
mà: $\widehat{ACB}$ = $\widehat{ACM}$ + $\widehat{BCM}$
⇒ $\widehat{ACM}$ = $\widehat{BCM}$ = $\cfrac{\widehat{ACB}}{2}$
⇒ $\widehat{ACM}$ = $\widehat{BCM}$ = $\cfrac{60^o }{2}$
= $30^o$
Xét ΔACM, có: $\widehat{MAC}$ + $\widehat{ACM}$ + $\widehat{AMC}$ = $180^o$
⇒ $\widehat{AMC}$ = $180^o$ - $(\widehat{MAC}$ + $\widehat{ACM})$
⇒ $\widehat{AMC}$ = $180^o$ - $(50^o$ + $30^o)$
= $180^o$ - $80^o$
= $100^o$
Xét ΔBCM, có: $\widehat{MBC}$ + $\widehat{BCM}$ + $\widehat{BMC}$ = $180^o$
⇒ $\widehat{BMC}$ = $180^o$ - $(\widehat{MBC}$ + $\widehat{BCM})$
⇒ $\widehat{BMC}$ = $180^o$ - $(70^o$ + $30^o)$
= $180^o$ - $100^o$
= $80^o$
Vậy: $\widehat{AMC}$ = $100^o$
$\widehat{BMC}$ = $80^o$
Chúc bn học tốt!
Mong bn cho mik ctlhn và vote 5* ah
