# Giải bài 1 trang 107 - Sách giáo khoa Toán 7 tập 1

##### Hướng dẫn giải

- Theo hình 47 ta có :

$$\triangle$$ABC có : $$\widehat{A}+\widehat{B}+\widehat{C}=180^0 \Rightarrow \widehat{C}=180^0-(\widehat{A}+\widehat{B})$$

$$\Rightarrow x = 180^0-(90^0+55^0)=35^0$$

- Theo hình 48 ta có :

$$\triangle$$GHI có : $$\widehat{G}+\widehat{H}+\widehat{I}=180^0 \Rightarrow \widehat{H}=180^0-(\widehat{G}+\widehat{I})$$

$$\Rightarrow x = 180^0-(30^0+40^0)=110^0$$

- Theo hình 49 ta có :

$$\triangle$$MNP có : $$\widehat{M}+\widehat{N}+\widehat{P}=180^0 \Rightarrow \widehat{M}+\widehat{P}=180^0-\widehat{N}$$

$$\Rightarrow 2x = 180^0-50^0=130^0 \Rightarrow x = \dfrac{130^0}{2}=65^0$$

- Theo hình 50 ta có :

$$\widehat{D}+\widehat{E}+\widehat{K}=180^0 \Rightarrow \widehat{D}=180^0-(\widehat{E}+\widehat{K})$$

$$\Rightarrow \widehat{D} = 180^0-(60^0+40^0)=80^0$$

Mặt khác:

$$y+\widehat{D}=180^0 \Rightarrow y = 180^0 - \widehat{D}=180^0-80^0=100^0$$

Vì x + $$40^0$$ = $$180^0 \Rightarrow x = 180^0 - 40^0=140^0$$

- Theo hình 51 ta có :

$$\widehat{ABD}+\widehat{BAD}+\widehat{ADB}=180^0 \Rightarrow \widehat{ADB}=180^0-(\widehat{ABD}+\widehat{BAD})$$

$$\Rightarrow \widehat{ADB} = 180^0-(70^0+40^0)=70^0$$

Mặt khác :

$$x + \widehat{ADB}=180^0 \Rightarrow x = 180^0-\widehat{ADB}$$

$$\Rightarrow x = 180^0 - 70^0 = 110^0$$

Ta có :

$$\widehat{ADC}+\widehat{DCA}+\widehat{CAD}= 180^0 \Rightarrow \widehat{DCA}=180^0-(\widehat{ADC}+\widehat{CAD})$$

$$\Rightarrow y = 180^0-110^0+40^0=30^0$$