a) Ta có :

`Â` + B + C = $180^{0}$

` B + C = $90^{0}$ `

`=>` B = ( 90 + 20 ) : 2 = $55^{0}$

C = 90 - 55 = $35^{0}$

b) Ta có :$Â_1$ + $Â_2$ = $180^{0}$

$B_1$ + $B_2$ = $180^{0}$

$C_1$ + $C_2$ = $180^{0}$

Theo vế :

$Â_1$ + $Â_2$ + $B_1$ + $B_2$ + $C_1$ + $C_2$ = 180 + 180 + 180 = $540^{0}$

`=>` $Â_1$ + $B_1$ + $C_1$ = $180^{0}$ ( tổng định lý  3 gốc trong 1 tam giác)

$Â_2$ + $Â_2$ + $Â_2$ = $360^{0}$

@tranlatuduy

 

 

Giải thích các bước giải:

`a)` Ta có: `\hat{A}+\hat{B}+\hat{C}=180^o
 => \hat{B}+\hat{C}=90^o`
Mà `\hat{B}-\hat{C}=20^o`
`=> \hat{B}=(90+20):2=55^o`; `\hat{C}=90-55=35^o`

`b)` Ta có: `\hat{A_1}+\hat{A_2}=180^o`

`\hat{B_1}+\hat{B_2}=180^o`

`\hat{C_1}+\hat{C_2}=180^o`

Cộng vế theo vế:  `\hat{A_1}+\hat{A_2}+\hat{B_1}+\hat{B_2}+\hat{C_1}+\hat{C_2}=180+180+180=540^o`

Mà `\hat{A_1}+\hat{B_1}+\hat{C_1}=180^o` (tổng `3` góc trog `1 Δ`)

`=> \hat{A_2}+\hat{A_2}+\hat{A_2}=360^o`