a: \(\widehat{A}=2\cdot\widehat{B}=3\cdot\widehat{C}\)
=>\(\widehat{B}=\dfrac{\widehat{A}}{2};\widehat{C}=\dfrac{\widehat{A}}{3}\)
Xét ΔABC có \(\widehat{A}+\widehat{B}+\widehat{C}=180^0\)
=>\(\widehat{A}+\dfrac{\widehat{A}}{2}+\dfrac{\widehat{A}}{3}=180^0\)
=>\(\dfrac{11}{6}\cdot\widehat{A}=180^0\)
=>\(\widehat{A}=180^0:\dfrac{11}{6}=180^0\cdot\dfrac{6}{11}=\dfrac{1080^0}{11}\)
\(\widehat{B}=\dfrac{1080^0}{11}:2=\dfrac{540^0}{11};\widehat{C}=\dfrac{1080^0}{11}:3=\dfrac{360^0}{11}\)
b: Xét ΔABC có \(\widehat{B}=\widehat{C}\)
nên ΔABC cân tại A
\(\widehat{A}-\widehat{B}=60^0\)
=>\(\widehat{A}=60^0+\widehat{B}\)
ΔBCA cân tại A
=>\(\widehat{A}=180^0-2\cdot\widehat{B}\)
=>\(60^0+\widehat{B}=180^0-2\cdot\widehat{B}\)
=>\(3\cdot\widehat{B}=180^0-60^0=120^0\)
=>\(\widehat{B}=\dfrac{120^0}{3}=40^0\)
\(\widehat{C}=\widehat{B}=40^0;\widehat{A}=60^0+40^0=100^0\)
