Question

Tam giác ABC có góc B - góc C = 30°, 2 góc C - góc A = 100. Tính góc A, B,C.

Answer 1

`@\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`

Answer 2

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:$3c-(a+b)=70^0$

$\Rightarrow 4c-(a+b+c)=70^0$

$\Rightarrow 4c-180^0=70^0$

$\Rightarrow c=62,5^0$

$a=2c-100^0=2.62,5^0-100^0=25^0$

$b=30^0+c=30^0+62,5^0=92,5^0$