Ta có:
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)(Tổng ba góc trong một tam giác)
\(\Rightarrow\widehat{B}+\widehat{C}=180^o-\widehat{A}\)
\(\Rightarrow\widehat{B}+\widehat{C}=180^o-60^o\)
\(\Rightarrow\widehat{B}+\widehat{C}=120^o\)
Mà \(B=\dfrac{1}{2}\widehat{C}\)
\(\Rightarrow\dfrac{1}{2}\widehat{C}+\widehat{C}=120^o\)
\(\Rightarrow\dfrac{3}{2}\widehat{C}=120^o\)
\(\Rightarrow\widehat{C}=80^o\)
\(\Rightarrow\widehat{B}=\dfrac{1}{2}\cdot\widehat{C}=\dfrac{1}{2}\cdot80^o=40^o\)
ゆきのよう
xét \(\Delta ABC\) có
\(\widehat{A}+\widehat{B}+\widehat{C}=180^o\) (tổng 3 góc trong tam giác)
\(=>60^o+\widehat{B}+\widehat{C}=180^o\\ =>\widehat{B}+\widehat{C}=120^o\)
mà \(\widehat{B}=\dfrac{1}{2}\widehat{C}=>\dfrac{\widehat{B}}{\widehat{C}}=\dfrac{1}{2}=>\dfrac{\widehat{B}}{1}=\dfrac{\widehat{C}}{2}\)
áp dụng tích chất dãy tỉ số bằng nhau ta có
\(\dfrac{\widehat{B}}{1}=\dfrac{\widehat{C}}{2}=\dfrac{\widehat{B}+\widehat{C}}{1+2}=\dfrac{120^o}{3}=40^o\\ =>\widehat{B}=40^o\cdot1=40^o\\ =>\widehat{C}=40^o\cdot2=80^o\)
