Toán 9

Y

yumi_26

Cho tam giác ABC có 3 góc nhọn, BC= a , AC = b , AB = c. CMR
[TEX]\frac{a}{SinA} = \frac{b}{SinB} = \frac{c}{SinC}[/TEX]

Vẽ AH [TEX] \perp \[/TEX] BC, H [TEX] \in [/TEX] BC;
[TEX] \triangle \ [/TEX] HAB có [TEX] \hat{H} = 90^o[/TEX] \Rightarrow [TEX] sinB = \frac{AH}{AB} [/TEX]
[TEX] \triangle \ [/TEX] AHC có [TEX] \hat{H} = 90^o[/TEX] \Rightarrow [TEX] sinC = \frac{AH}{AC} [/TEX]
\Rightarrow [TEX] \frac{sinB}{sinC} = \frac{AC}{AB} = \frac{b}{c}[/TEX] \Rightarrow [TEX] \frac{b}{sinB} = \frac{c}{sinC}[/TEX]
Chứng minh tương tự ta có: [TEX] \frac{a}{sinA} = \frac{b}{sinB}[/TEX]
Vậy [TEX] \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC}[/TEX]
 
Q

quynhnhung81

Kẻ CH [TEX]\perp[/TEX]AB
Ta có [TEX]sin A=\frac{CH}{b} \ va \ sin B = \frac{CH}{a}[/TEX]

[TEX]\Rightarrow \frac{sin A}{sin B}=\frac{\frac{CH}{b}}{\frac{CH}{a}}=\frac{a}{b}[/TEX]

[TEX]\Rightarrow \frac{a}{sinA}=\frac{b}{sin B} \ \ \ \ \ \ \ \ \ (1)[/TEX]
kẻ AK[TEX]\perp[/TEX]BC
ta có [TEX]sin C=\frac{AK}{b} \ va \ sin B = \frac{AK}{c}[/TEX]

[TEX]\Rightarrow \frac{sin C}{sin B}=\frac{\frac{AK}{b}}{\frac{AK}{c}}=\frac{c}{b}[/TEX]

[TEX]\Rightarrow \frac{c}{sinC}=\frac{b}{sin B} \ \ \ \ \ \ \ \ \ (2)[/TEX]

từ (1) và (2) suy ra dpcm
 
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