`4A)`
`\triangleABC` `có:`
`\hat{A}+\hat{B}+\hat{C}=180^o`
`a)\frac{\hat{A}}{6}=\frac{\hat{B}}{3}=\frac{\hat{C}}{1}`
`\text{Áp dụng tính chất dãy tỉ số bằng nhau, ta có:}`
`\frac{\hat{A}}{6}=\frac{\hat{B}}{3}=\frac{\hat{C}}{1}=\frac{\hat{A}+\hat{B}+\hat{C}}{6+3+1}=\frac{180^o}{10}=18^o`
`Do` `đó:`
`@\hat{A}=18^o .6=108^o`
`@\hat{B}=18^o .3=54^o`
`@\hat{C}=18^o .1=18^o`
`b)\frac{\hat{A}}{2}=\frac{\hat{B}}{3}=\frac{\hat{C}}{4}`
`\text{Áp dụng tính chất dãy tỉ số bằng nhau, ta có:}`
`\frac{\hat{A}}{2}=\frac{\hat{B}}{3}=\frac{\hat{C}}{4}=\frac{\hat{A}+\hat{B}+\hat{C}}{2+3+4}=\frac{180^o}{9}=20^o`
`Do` `đó:`
`@\hat{A}=20^o .2=40^o`
`@\hat{B}=20^o .3=60^o`
`@\hat{C}=20^o .4=80^o`
`4B)`
`a)\hat{C}-\hat{B}=36^o`
`<=>\hat{C}=\hat{B}+36^o`
`Ta` `có:`
`\hat{A}+\hat{B}+\hat{C}=180^o`
`<=>2\hat{B}+\hat{B}+\hat{B}+36^o=180^o`
`<=>4\hat{B}=144^o`
`<=>\hat{B}=36^o`
`=>\hat{A}=2\hat{B}=72^o`
`\hat{C}=36^o +\hat{B}=72^o`
`b)\text{Áp dụng tính chất dãy tỉ số bằng nhau, ta có:}`
`\frac{\hat{A}}{3}=\frac{\hat{B}}{1}=\frac{\hat{C}}{2}=\frac{\hat{A}+\hat{B}+\hat{C}}{3+1+2}=\frac{180^o}{6}=30^o`
`Do` `đó:`
`@\hat{A}=30^o .3=90^o`
`@\hat{B}=30^o .1=30^o`
`@\hat{C}=30^o .2=60^o`
