Đặt \(\widehat{A}=a;\widehat{B}=b;\widehat{C}=c\)
Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
=>\(a+b+c=180^0\)
a: \(\widehat{C}-\widehat{B}=36^0\)
=>\(c-b=36^0\)
=>\(c=b+36^0\)
\(\widehat{A}=2\cdot\widehat{B}\)
=>a=2b
\(a+b+c=180^0\)
=>\(2b+b+b+36^0=180^0\)
=>\(4b=144^0\)
=>\(b=36^0\)
\(\widehat{B}=b=36^0\)
\(\widehat{A}=2\cdot36^0=72^0\)
\(\widehat{C}=36^0+36^0=72^0\)
b: \(\dfrac{\widehat{A}}{3}=\dfrac{\widehat{B}}{1}=\dfrac{\widehat{C}}{2}\)
=>\(\dfrac{a}{3}=\dfrac{b}{1}=\dfrac{c}{2}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{3}=\dfrac{b}{1}=\dfrac{c}{2}=\dfrac{a+b+c}{3+1+2}=\dfrac{180^0}{6}=30^0\)
=>\(a=3\cdot30^0=90^0;b=1\cdot30^0=30^0;c=2\cdot30^0=60^0\)
\(\widehat{A}=a=90^0;\widehat{B}=b=30^0;\widehat{C}=c=60^0\)
