`\Delta ABC = \Delta MNP`

`=> \hat{A} = \hat{M} = 40^o`

Từ GT: `3\hat{B} = 4\hat{C}`

`=> \hat{B} = 4/3 \hat{C}`

Xét `\Delta ABC` ta có:

`\hat{A} + \hat{B} + \hat{C} = 180^o`

`=> 40^o + 4/3 \hat{C} + \hat{C} = 180^o`

`=> 7/3 \hat{C} = 140^o`

`=> \hat{C} = 60^o`

`=> \hat{B} = 4/3 . 60^o = 80^o`

Vậy `\hat{A} = 40^o ; \hat{B} = 80^o ; \hat{C} = 60^o`

`@` tremamnon

Đáp án:

Vì `ΔABC=ΔMNP`

`to \hatA=\hatM=40^o`

Ta có :

`3\hatB=4\hatC`

`to (\hatB)/4=(\hatC)/3`

`to (\hatB)/4*3=\hatC`

`to 3/4*\hatB=\hatC`

Xét `ΔABC` có :

`\hatA+\hatB+\hatC=180^o` ( tổng các góc trong tam giác )

`to 40^{o}+\hatB+3/4\hatB=180^o`

`to 7/4\hatB=140^o`

`to \hatB=80^o`

`to \hatC=\hatB:4/3=80^o:4/3=60^o`

Vậy : `\hatA=40^o;\hatB=80^o;\hatC=60^o`