Question

tui cần gấp hôm nay ạ . tui cảm ơn ạ

Answer 1

Xét tứ giác ABCD có \(\widehat{BAD}+\widehat{ADC}+\widehat{ABC}+\widehat{BCD}=360^0\)

=>\(\widehat{ABC}+\widehat{BCD}+70^0+50^0=360^0\)

=>\(\widehat{ABC}+\widehat{BCD}=240^0\)

=>\(\widehat{MBC}+\widehat{MCB}=\dfrac{240^0}{2}=120^0\)

Xét ΔMBC có \(\widehat{MBC}+\widehat{MCB}+\widehat{BMC}=180^0\)

=>\(\widehat{BMC}+120^0=180^0\)

=>\(\widehat{BMC}=60^0\)