Question

[Vũ Xuân Huy]

cứu mik với, mik cần gấp

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Ta có: \(\widehat{bMB}=\widehat{NMC}\)(hai góc đối đỉnh)

mà \(\widehat{bMB}=50^0\)

nên \(\widehat{NMC}=50^0\)

Ta có: \(\widehat{MNC}+\widehat{aNC}=180^0\)(hai góc kề bù)

=>\(\widehat{MNC}+110^0=180^0\)

=>\(\widehat{MNC}=70^0\)

Xét ΔMNC có \(\widehat{NMC}+\widehat{MNC}+\widehat{C}=180^0\)

=>\(\widehat{C}+50^0+70^0=180^0\)

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

b: Ta có: \(\widehat{NMB}+\widehat{NMC}=180^0\)(hai góc kề bù)

=>\(\widehat{NMB}+50^0=180^0\)

=>\(\widehat{NMB}=130^0\)

Ta có: MN//AB

=>\(\widehat{CMN}=\widehat{CBA}\)(hai góc đồng vị)

=>\(\widehat{CBA}=50^0\)

BN là phân giác của góc CBA

=>\(\widehat{NBM}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔNMB có \(\widehat{NMB}+\widehat{BNM}+\widehat{NBM}=180^0\)

=>\(\widehat{MNB}=180^0-130^0-25^0=25^0\)

c: BN là phân giác của góc CBA

=>\(\widehat{ABN}=\dfrac{\widehat{ABC}}{2}=25^0\)

Xét ΔABC có \(\widehat{ABC}+\widehat{ACB}+\widehat{BAC}=180^0\)

=>\(\widehat{BAN}+60^0+50^0=180^0\)

=>\(\widehat{BAN}=70^0\)

Xét ΔBAN có \(\widehat{BAN}+\widehat{ABN}+\widehat{ANB}=180^0\)

=>\(\widehat{ANB}=180^0-75^0-25^0=85^0\)