Question

Cho tam giác ABC. Tính số đo \(\widehat{A}\) , \(\widehat{B}\) , \(\widehat{C}\), biết:

a, \(\widehat{A}\) = 80 độ, \(\widehat{B}\) - \(\widehat{C}\) = 20 độ

b, \(\widehat{A}\) =70 độ; \(\widehat{B}\)= 3\(\widehat{C}\)

Answer 1

a) ta có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{B}+\widehat{C}=100^0\Leftrightarrow\widehat{B}=100^0-\widehat{C}\)

mà \(\widehat{B}-\widehat{C}=20^0\Leftrightarrow100^0-\widehat{C}-\widehat{C}=20^0\Leftrightarrow\widehat{C}=40^0\)

vậy \(\widehat{B}=100^0-\widehat{C}=60^0\)

b) ta có \(\widehat{B}=3\widehat{C}\)

mà \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\Leftrightarrow\widehat{B}+\widehat{C}=110^0\Leftrightarrow4\widehat{C}=110^0\Rightarrow\widehat{C}=27,5^0\)

\(\widehat{B}=3\widehat{C}=27,5^0.3=82,5^0\)