` Bài 1:`

` Hình 1:`

`cos20^0=(BH)/(AH)`

`=>x=20:cos20^0`

`=>` `x ` `~~` `21,28`

` triangleAHB` ` vuông `  ` tại ` ` H`

`=>AH^2=AB^2-BH^2`

`=>` `AH` `~~` `  7,27`

`cos45^0=sqrt{2}/2`

`=>(AH)/y=sqrt{2}/2`

`=>y ` `~~``10,28`

` Vậy ` `x ` `~~` `21,28` `,` `y ` `~~` `10,28`

`Hình 2:` `s``i``n``40^0=y/(KE)`

`=>y=30.``s``i``n``40^0`

`=>y `  `~~` `19,28`

`s``i``n``60^0=sqrt{3}/2`

`=>y/x=sqrt{3}/2`

`=>x ` `~~` `22,26`

` Vậy ` `x ` `~~` `22,26``,` `y `  `~~` `19,28`

` Bài ` `2:`

`a,cos``hat{A}=(AH)/(AB)`

`cos hat{A}=(AK)/(AC)`

`=>(AK)/(AC)=(AH)/(AB)`

`=>AK.AB=AC.AH`

` Vậy ` `AK.AB=AC.AH`

`b,` 

` Cách 1` ` Dùng ` ` ~`

`triangleAHK`  ` và `  ` triangle ABC` ` có:`

`hat{A} ` ` chung`

`(AK)/(AC)=(AH)/(AB)` `( câu ` ` a)`

`=>` `triangleAHK`  ` ~ `  ` triangle ABC`

`=>k=(AH)/(AB)`

`=>(S_(AHK))/(S_(ABC))=(AH^2)/(AB^2)`

` Mà ` ` cos45^0=(AH)/(AB)=sqrt{2}/2`

`=>(AH^2)/(AB^2)=1/2`

`=>S_(AHK)=1/2 S_(ABC)`

` Mà `  `S_(AHK)+S_(BCHK)=S_(ABC)`

`=>S_(BCHK)=1/2 S_(ABC) =S_(AHK)`

` Vậy ` ` S_(AHK)=S_(BCHK)`

` Cách 2:`

`S_(AHK)=AK.AH.` `s``i``n``hat{A}`

`=AK.AH.``s``i``n``45^0`

`=sqrt{2}/2 . AK.AH`

`S_(ABC)=AB.AC.``s``i``n``hat{A}`

`=AB.AC . sqrt{2}/2`

` Mà ` ` cos45^0=(AH)/(AB)=(AK)/(AC)=sqrt{2}/2`

`=>AH.AC=1/2 . AB.AC`

`=>S_(AHK)=1/2 . S_(ABC)`

` Mà `  `S_(AHK)+S_(BCHK)=S_(ABC)`

`=>S_(BCHK)=1/2 S_(ABC) =S_(AHK)`

` Vậy ` ` S_(AHK)=S_(BCHK)`

Bài 1:

Ta có: `\mathbb{AB} = \mathbb{BH}/(cos\mathbb{B})` (hệ thức trong tam giác vuông)

`=> \mathbb{AB} = 20/(cos 20^\text{o}) ~~ 21,28` hay `x = 21,28`

Áp dụng định lí Py-ta-go vào `\mathbb{ΔABH}` ta được:

`\mathbb{AB}^2 = \mathbb{AH}^2 + \mathbb{BH}^2` `=> \mathbb{AH}^2 = \mathbb{AB}^2 - \mathbb{BH}^2`

Hay `\mathbb{AH}^2 = 21,28^2 - 20^2 = 52,8384` `=> \mathbb{AH} = \sqrt{52,8384}`

Xét `\mathbb{ΔAHC}` có: `\mathbb{AC} = \mathbb{AH}/(cos\mathbb{A})` (định lí)

`=> \mathbb{AC} = (\sqrt{52,8384})/(cos 45^\text{o}) ~~ 10,28` hay `y = 10,28`

Bài 2:

`\text{a)}` Xét `\mathbb{ΔABK}` có: `cos \mathbb{A} = \mathbb{AK}/\mathbb{AB}`

Xét `\mathbb{ΔACH}` có: `cos \mathbb{A} = \mathbb{AH}/\mathbb{AC}`

Do đó: `\mathbb{AK}/\mathbb{AB} = \mathbb{AH}/\mathbb{AC}` hay `\mathbb{AH. AC} = \mathbb{AK. AB}`

`\text{b)}` Xét `\mathbb{ΔAHK}` và `\mathbb{ΔABC}` ta có:

                `\mathbb{AK}/\mathbb{AB} = \mathbb{AH}/\mathbb{AC}` (cmt)

                `\hat{\mathbb{A}}` là góc chung

`=> \mathbb{ΔAHK}` $\backsim$ `\mathbb{ΔABC}` (c . g . c)

Do đó: `k = \mathbb{AH}/\mathbb{AB}` `=> \bb{S}_\mathbb{AHK}/\bb{S}_\mathbb{ABC} = (\mathbb{AH}/\mathbb{AB})^2`

Mà `cos \mathbb{A} = \mathbb{AH}/\mathbb{AB} = cos 45^\text{o} = \sqrt{2}/2`

`=> \bb{S}_\mathbb{AHK} = \bb{S}_\mathbb{ABC} . (\sqrt{2}/2)^2 = 1/2. \bb{S}_\mathbb{ABC}` `(1)`

Mặt khác: `\bb{S}_\mathbb{BCHK} = \bb{S}_\mathbb{ABC} - \bb{S}_\mathbb{AHK}`

Hay `\bb{S}_\mathbb{ABC} - 1/2 \bb{S}_\mathbb{ABC} = 1/2 \bb{S}_\mathbb{ABC}` `(2)`

Từ `(1)` và `(2)` ta suy ra: `\bb{S}_\mathbb{AHK} = \bb{S}_\mathbb{BCHK}`