Bài 1:ADHTL vào ΔABC($\widehat{A}$=$90^o$) có:

$AB^{2}$ =BC.BH

$\Leftrightarrow$$9^{2}$=15.BH

$\Rightarrow$BH=$9^{2}$:15=81:15=5,4(cm)

Ta có:BC=CH+BH

$\Leftrightarrow$15=CH+5,4

$\Rightarrow$CH=15-5,4=9,6(cm)

ADHTL vào ΔABC($\widehat{A}$=$90^o$) có:

$AC^{2}$ =BC.CH

$\Leftrightarrow$$AC^{2}$=15.9,6=144

$\Leftrightarrow$AC=$\sqrt[]{144}$ =12(cm)

ADHTL vào ΔABC($\widehat{A}$=$90^o$) có:

AB.AC=BC.AH

$\Leftrightarrow$9.12=15.AH

$\Leftrightarrow$108=15.AH

$\Rightarrow$AH=108:15=7,2(cm)

Bài 2:

Ta có:$\widehat{B}$+$\widehat{C}$=$90^o$

$\Rightarrow$$\widehat{B}$=$90^o$-$\widehat{C}$=$90^o$-$30^o$=$60^o$

ADTSLG vào ΔABC($\widehat{A}$=$90^o$) ta có:

cos C=$\frac{k}{h}$ =$\frac{AC}{BC}$

$\Leftrightarrow$ cos $30^o$=$\frac{AC}{10}$

$\Rightarrow$AC=cos $30^o$.10= $\frac{\sqrt[]{3}}{2}$ .10=5$\sqrt[]{3}$ (cm)

sin C=$\frac{đ}{h}$ =$\frac{AB}{BC}$

$\Leftrightarrow$ sin $30^o$=$\frac{AB}{10}$

$\Rightarrow$AB=sin $30^o$.10= $\frac{1}{2}$ .10=5(cm)

Đáp án: 

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