Question

cho tam giác ABC (góc B<góc C) có góc A+2B=100 độ. tính số đo góc C-B

Answer 1

Ta có: \(\widehat{A}+2.\widehat{B}=100^0\) => \(\widehat{A}=100^0-2.\widehat{B}\)

Xét t/giác ABC có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)(tổng 3 góc của 1 t/giác)

=> \(\left(100^0-2.\widehat{B}\right)+\widehat{B}+\widehat{C}=180^0\)

=> \(100^0-2\widehat{B}+\widehat{B}+\widehat{C}=180^0\)

=> \(100^0-\widehat{B}+\widehat{C}=180^0\)

=> \(\widehat{C}-\widehat{B}=180^0-100^0\)

=> \(\widehat{C}-\widehat{B}=80^0\)

Answer 2

cảm ơn nhiều