$\widehat{AEF}+\widehat{AEy}=180^o$ (2 góc kề bù)

$⇒\widehat{AEF}+128^o=180^o$

$⇒\widehat{AEF}=180^o-128^o$

$⇒\widehat{AEF}=52^o$

Xét $ΔAEF$ có :

$\widehat{AEF}+\widehat{AFE}+\widehat{BAC}=180^o$

$⇒52^o+46^o+\widehat{BAC}=180^o$

$⇒98^o+\widehat{BAC}=180^o$

$⇒\widehat{BAC}=180^o-98^o$

$⇒\widehat{BAC}=82^o$

Vậy $\widehat{BAC}=82^o$

Xét $ΔABC$ có:

$\widehat{ABC}+\widehat{BAC}+\widehat{ACB}=180^o$

$⇒52^o+82^o+\widehat{ACB}=180^o$

$⇒134^o+\widehat{ACB}=180^o$

$⇒\widehat{ACB}=180^o-134^o$

$⇒\widehat{ACB}=46^o$

Vậy $\widehat{ACB}=46^o$

2. Tính ∠BAC 

   Vì xx'//yy' ⇒ ∠yEA = ∠x'AE = $128^{0}$ 

                   ⇒ ∠x'AF = ∠yFA = $46^{0}$ ( gt)

    ⇒ ∠BAC = ∠EAx' - ∠FAx' = $128^{0}$ - $46^{0}$

   ⇒ ∠BAC = $82^{0}$ 

Tính ∠ACB

  ∠AFy = ∠y'FC = $46^{0}$ ( đối đỉnh )

Mà 2 góc y'FC và ACB nằm ở vị trí so le trong

  ⇒ ∠y'FC = ∠ACB = $46^{0}$

@Sứa

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