Question

Answer 1

\(\widehat{C}+\widehat{A}=180^0-60^0=120^0\)

mà \(\widehat{C}< \widehat{A}\)

nên \(\widehat{C}< 60^0;\widehat{A}>60^0\)

Xét ΔABC có \(\widehat{C}< \widehat{B}< \widehat{A}\)

nên AB<AC<BC

Answer 2

Xét `△ABC` có:

 \(\widehat{B}=60^0\)

mà \(\widehat{C}< \widehat{A}\)

nên

 \(\widehat{A}+\widehat{C}=180^0-\widehat{B}\)

\(\widehat{A}+\widehat{C}=180^0-60^0\)

\(\widehat{A}+\widehat{C}=120^0\)

Mà: \(\widehat{C}< \widehat{A}\)

`=>` \(60^0< \widehat{A}< 120^0\)

\(\widehat{C}< 60^0\)

Trong `△ABC`  có:

\(\widehat{C}< \widehat{B}< \widehat{A}\)

`=> AB < AC < BC`