`@\hat{B}-\hat{C}=30^o =>\hat{B}=30^o +\hat{C}`
`@2\hat{C}-\hat{A}=100^o =>\hat{A}=2\hat{C}-100^o`
Xét `\triangle ABC` có: `\hat{A}+\hat{B}+\hat{C}=180^o`
`=>2\hat{C}-100^o +30^o +\hat{C}+\hat{C}=180^o`
`=>4\hat{C}=250^o`
`=>\hat{C}=62,5^o`
`=>\hat{B}=30^o +62,5^o =92,5^o`
và `\hat{A}=2.62,5^o -100^o =25^o`
Lời giải:
Gọi độ đo góc $\widehat{A}, \widehat{B}, \widehat{C}$ lần lượt là $a,b,c$. Ta có:
$b-c=30^0(1)$
$2c-a=100^0(2)$
$a+b+c=180^0(3)$
Lấy $(2)-(1)$ có:$3c-(a+b)=70^0$
$\Rightarrow 4c-(a+b+c)=70^0$
$\Rightarrow 4c-180^0=70^0$
$\Rightarrow c=62,5^0$
$a=2c-100^0=2.62,5^0-100^0=25^0$
$b=30^0+c=30^0+62,5^0=92,5^0$