Giải bài 53 trang 89 – SGK Toán lớp 9 tập 2
Biết ABCD là tứ giác nội tiếp. Hãy điền vào ô trống trong bảng sau (nếu có thể ):
| Góc | 1) | 2) | 3) | 4) | 5) | 6) |
| \(\widehat{A}\) | \(80^o\) | \(60^o\) | \(95^o\) | |||
| \(\widehat{B}\) | \(70^o\) | \(40^o\) | \(65^o\) | |||
| \(\widehat{C}\) | \(105^o\) | \(74^o\) | ||||
| \(\widehat{D}\) | \(75^o\) | \(98^o\) |
Hướng dẫn:
Sử dụng định lý: "Trong một tứ giác nội tiếp, tổng số đo hai góc đối nhau bằng \(180^o\)"
TH1: \(\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{80}^{o}}+\widehat{C}={{180}^{o}} \\ & {{70}^{o}}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{C}={{100}^{o}} \\ & \widehat{D}={{110}^{o}} \\ \end{aligned} \right. \)
TH2: \(\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}+{{105}^{o}}={{180}^{o}} \\ & \widehat{B}+{{75}^{o}}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}={{75}^{o}} \\ & \widehat{B}={{105}^{o}} \\ \end{aligned} \right. \)
TH3: \(\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & 60^o+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{C}=120^o \\ & \widehat{B}=\alpha \\ &\widehat{D}=180^o-\alpha \end{aligned} \right. \)
TH4: \(\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ &40^o+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}=\alpha \\ & \widehat{C}=180^o-\alpha \\ &\widehat{D}=140^o \end{aligned} \right. \)
TH5: \(\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}+{{74}^{o}}={{180}^{o}} \\ & \widehat{D}+{{65}^{o}}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{A}={{106}^{o}} \\ & \widehat{D}={{115}^{o}} \\ \end{aligned} \right. \)
TH6: \(\left\{ \begin{aligned} & \widehat{A}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+\widehat{D}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & {{95}^{o}}+\widehat{C}={{180}^{o}} \\ & \widehat{B}+{{98}^{o}}={{180}^{o}} \\ \end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned} & \widehat{C}={{85}^{o}} \\ & \widehat{B}={{82}^{o}} \\ \end{aligned} \right. \)
Từ đó, ta có kết quả trong bảng sau:
| Góc | 1) | 2) | 3) | 4) | 5) | 6) |
| \(\widehat{A}\) | \(80^o\) | \(75^o\) | \(60^o\) | \(0<\alpha<180^o\) | \(106^o\) | \(95^o\) |
| \(\widehat{B}\) | \(70^o\) | \(105^o\) | \(0<\alpha<180^o\) | \(40^o\) | \(65^o\) | \(82^o\) |
| \(\widehat{C}\) | \(100^o\) | \(105^o\) | \(120^o\) | \(180^o-\alpha\) | \(74^o\) | \(85^o\) |
| \(\widehat{D}\) | \(110^o\) | \(75^o\) | \(180^o-\alpha\) | \(140^o\) | \(115^o\) | \(98^o\) |