Cho(O) lấy 3 điểm A,B,C trên đường tròn sao cho góc BAC=45° góc ABC=60° so sánh các cung AB BC CA 24/10/2021 Bởi Hailey Cho(O) lấy 3 điểm A,B,C trên đường tròn sao cho góc BAC=45° góc ABC=60° so sánh các cung AB BC CA
Xét $∆ABC$ có: `\hat{BAC}+\hat{ABC}+\hat{ACB}=180°` (tổng $3$ góc trong ∆ bằng $180°$) `=>\hat{ACB}=180°-(\hat{BAC}+\hat{ABC})` `=180°-(45°+60°)=75°` Vì `45°<60°<75°` `=>\hat{BAC}<\hat{ABC}<\hat{ACB}` Mà: `\hat{BAC}=1/ 2 sđ\stackrel\frown{BC}` (góc nội tiếp chắn cung $BC$) `\hat{ABC}=1/ 2 sđ\stackrel\frown{CA}` (góc nội tiếp chắn cung $CA$) `\hat{ACB}=1/ 2 sđ\stackrel\frown{AB}` (góc nội tiếp chắn cung $AB$) `=>sđ\stackrel\frown{BC}<sđ\stackrel\frown{CA}<sđ\stackrel\frown{AB}` `=>\stackrel\frown{BC}<\stackrel\frown{CA}<\stackrel\frown{AB}` Bình luận
$\begin{cases}\widehat{A}=45{}^\circ\\\\\widehat{B}=60{}^\circ\end{cases}\,\,\,\to\,\,\,\widehat{C}=180{}^\circ-\left(45{}^\circ+60{}^\circ\right)\,\,\,\to\,\,\,\widehat{C}=75{}^\circ$ $\,\,\,\,\,\,\widehat{A}\,\,<\,\,\widehat{B}\,\,<\,\,\widehat{C}\,\,\left( \,\,45{}^\circ \,\,<\,\,60{}^\circ \,\,<\,\,75{}^\circ \,\, \right)$ $\to \overset\frown{BC}\,\,<\,\,\overset\frown{CA}\,\,<\,\,\overset\frown{AB}$ Bình luận
Xét $∆ABC$ có:
`\hat{BAC}+\hat{ABC}+\hat{ACB}=180°` (tổng $3$ góc trong ∆ bằng $180°$)
`=>\hat{ACB}=180°-(\hat{BAC}+\hat{ABC})`
`=180°-(45°+60°)=75°`
Vì `45°<60°<75°`
`=>\hat{BAC}<\hat{ABC}<\hat{ACB}`
Mà:
`\hat{BAC}=1/ 2 sđ\stackrel\frown{BC}` (góc nội tiếp chắn cung $BC$)
`\hat{ABC}=1/ 2 sđ\stackrel\frown{CA}` (góc nội tiếp chắn cung $CA$)
`\hat{ACB}=1/ 2 sđ\stackrel\frown{AB}` (góc nội tiếp chắn cung $AB$)
`=>sđ\stackrel\frown{BC}<sđ\stackrel\frown{CA}<sđ\stackrel\frown{AB}`
`=>\stackrel\frown{BC}<\stackrel\frown{CA}<\stackrel\frown{AB}`
$\begin{cases}\widehat{A}=45{}^\circ\\\\\widehat{B}=60{}^\circ\end{cases}\,\,\,\to\,\,\,\widehat{C}=180{}^\circ-\left(45{}^\circ+60{}^\circ\right)\,\,\,\to\,\,\,\widehat{C}=75{}^\circ$
$\,\,\,\,\,\,\widehat{A}\,\,<\,\,\widehat{B}\,\,<\,\,\widehat{C}\,\,\left( \,\,45{}^\circ \,\,<\,\,60{}^\circ \,\,<\,\,75{}^\circ \,\, \right)$
$\to \overset\frown{BC}\,\,<\,\,\overset\frown{CA}\,\,<\,\,\overset\frown{AB}$