he thuc luong trong tam giac

H

hermes_legend

Hình tự vẽ anh nhé :D. Kẻ thêm CH vuông góc AB.
Xét [tex]\large\Delta[/tex]ACH: CH=AC.sin [TEX]\{CAH}\Rightarrow[/TEX] CH=3,464.
AH=AC.cos[TEX]\{CAH}\Rightarrow[/TEX] AH=2(*)
Xét[tex]\large\Delta[/tex]BCH: [TEX]BC=\frac{CH}{sin \{CBH}}\Rightarrow[/TEX] BC=4,9.
BH=CH.cot [TEX]\{HBC}\Rightarrow[/TEX] BH=3,464.(**)
Từ (*)(**) [TEX]\Rightarrow[/TEX] AB=c=2+3,464=5,464
:D
 
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