^B+^C=1800-1000=800
=> ^C=(800-500)/2=150
^B=150+500=650
ĐS: ^B=650; ^C=150.
Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) và \(\widehat{A}=100^o\) ; \(\widehat{B}-\widehat{C}=50^o\)
\(\Rightarrow\widehat{A}+\widehat{B}+\widehat{C}-\widehat{A}=180^o-100^o\)
\(\Rightarrow\widehat{B}+\widehat{C}=80^0\) mà \(\widehat{B}-\widehat{C}=50^0\)
\(\Rightarrow\widehat{B}=\left(\widehat{B}+\widehat{C}-\left(\widehat{B}-\widehat{C}\right)\right)\left(80^o+50^0\right):2=65^0\)
\(\Rightarrow\widehat{C}=\widehat{B}-50^0=65^0-50^0=15^0\)
tam giác ABC có \(\widehat{A}\)+ \(\widehat{B}\)+ \(\widehat{C}\)= \(180^0\)
\(100^0\)+ \(\widehat{B}\)+ \(\widehat{C}\)= \(180^0\)
\(\Rightarrow\)\(\widehat{B}\)+ \(\widehat{C}\) = \(80^0\) (1)
mà \(\widehat{B}\)- \(\widehat{C}\)= \(50^0\)\(\Rightarrow\)\(\widehat{B}\)= \(50^0\)+ \(\widehat{C}\) (2)
Thay (2) vào (1) ta được:
\(50^0\)+ \(\widehat{C}\)+ \(\widehat{C}\)= \(180^0\)
\(50^0\)+ 2\(\widehat{C}\) = \(180^0\)
2\(\widehat{C}\) = \(130^0\)
\(\widehat{C}\) = \(65^0\)
\(\Rightarrow\) \(\widehat{B}\)= \(50^0\)+ \(65^0\)= \(115^0\)
Vậy...
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\Leftrightarrow100^o+\widehat{B}+\widehat{C}=180^o\Leftrightarrow\widehat{B}+\widehat{C}=80^o\)kết hợp giả thiết đề bài suy ra :\(\widehat{B}+\widehat{C}+\widehat{B}-\widehat{C}=80^o+50^o\Leftrightarrow2\widehat{B}=130^o\Leftrightarrow\widehat{B}=65^o\Leftrightarrow\widehat{C}=65^o-50^o=15^o\)
mik sửa lại đoạn này:
\(\Rightarrow\)\(\widehat{B}\)= \(65^0\)- \(50^0\)= \(15^0\)
MIK NHẦM CHÚT ĐỪNG ĐỂ Ý NHA
hình tự vẽ nhá
ta có: tổng 3 góc trg 1 tgiac bằng 180* (định lí tổng 3 góc trg 1 tam giác)
=> \(\widehat{A}\)+\(\widehat{B}\)+\(\widehat{C}\)= 180* mà \(\widehat{A}\)=100* => 100* +\(\widehat{B}\) + \(\widehat{C}\)=180* => góc B + góc C = 180* - 100* = 80*
=> góc B=( 80* + 50*) :2 = 65* ; góc C=(80* - 50*) : 2 = 15*
(*) dấu * ở đây mk thay cho kí hiệu của độ nha
