Ta có: `\sqrt{A_1}=120^o`(đối đỉnh)
`\hat{B_2}=55^o`(đối đỉnh)
`\hat{B_1}+55^o=180^o`
`<=>\hat{B_1}=180^o-55^o=135^o`
Vậy `\hat{A_1}=120^o`
`\hat{B_1}=135^o`
`\hat{B_2}=55^o`
Đáp án+Giải thích các bước giải:
Ta có: `\hat{A_1}= \hat{A_2}=120^0` (2 góc đối đỉnh)
Lại có: `\hat{B_2}= \hat{B_4}=55^0` (2 góc đối đỉnh)
Mặc khác ta có: `\hat{B_2}+\hat{B_3}=180^0` (2 góc kề bù)
`=> 55^0+\hat{B_3}=180^0`
`=> \hat{B_3}= 180^0-55^0= 125^0`
$@Maruko-chan$
Chúc mừng sinh nhật hoidap 🎉🎂 :3 Vẽ anime nữ và calli chữ "Chúc mừng sinh nhật hoidap 2 tuổi" nha :3 Xấu=BC Nl: Đang ngồi xem live :>
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