(tự vẽ hình)
a) Ta có: A^ +B^+C^ = 180o
B^ + C^ = 180o - 100o = 80o
b) MBA^ = MBC^ = ABC^/2
MCA^ = MCB^ = ACB^/2
Tam giác BMC:
MBC^ + MCB^ + BMC^ = 180o
ABC^/2 + ACB^/2 + BMC^ = 180o
40o + BMC^ = 180o
BMC^ = 140o
a, Xét \(\Delta ABC\) có
\(\widehat{A}+\widehat{B}+\widehat{C}=180\)*
Thay \(\widehat{A}=100*\)ta được
\(\widehat{B}+\widehat{C}=100*\)
b, Có \(\widehat{B}+\widehat{C}=100\text{*}\)
=> \(\frac{1}{2}\widehat{B}+\frac{1}{2}\widehat{C}=50^{\text{*}}\)
=> \(\widehat{MBC}+\widehat{MCB}=50^{\text{*}}\)
Xét \(\Delta MBC\) có
\(\widehat{MBC}+\widehat{MCB}=50^{\text{*}}\)
=> \(\widehat{BMC}=180^{\text{*}}-50^{\text{*}}=130^{\text{*}}\)
Sorry mình làm nhầm chỗ \(\widehat{B}+\widehat{C}\)
