Bài 1)
Áp dụng định lí Py-ta-go ta tính được $AC=√(BC²-AB²)=4^{}$ (cm)
$\frac{1}{AH^2}$ =$\frac{1}{AB^2}$ +$\frac{1}{AC^2}$
⇒ $AH=2,4^{}$ (cm)
Theo Py-ta-go ta tính được: $BH=√(AB²-AH²)=1,8^{}$ (cm)
$⇒ CH=BC-BH=5-1,8=3,2^{}$ (cm)
Bài 2)
$BC=√(AB²+AC²)≈12,8^{}$ (cm)
$\frac{1}{AH^2}$ =$\frac{1}{AB^2}$ +$\frac{1}{AC^2}$ ⇒ $AH≈6,25^{}$ (cm)
$BH=√(AB²-AH²)≈5^{}$ (cm)
$CH=BC-BH=12,8-5≈7,8^{}$ (cm)
Bài 3)
Kẻ $AH⊥BC^{}$
$⇒ AH=35.sin50≈26,8^{}$ (cm)
$HC=35.cos50≈22,5^{}$ (cm)
$BH=AH.cot60=26,8.0,58≈15,5^{}$ (cm)
$⇒ BC=BH+CH≈38^{}$ (cm)
$SΔABC=^{}$ $\frac{1}{2}$$. AH.BC=^{}$ $\frac{1}{2}$$.26,8.38≈1018,4^{}$ (cm²)
Bài 4)
a) $AH²=BH.CH ⇒ CH=^{}$ $\frac{AH^2}{BH}$=$\frac{16^2}{25}$=10,24
$BC=BH+CH=25+10,24=35,24 ^{}$
$AC=√(AH²+CH²)≈19 ^{}$
$AB=√(BC²-AC²)≈29,7 ^{}$
b) $SΔABC=^{}$ $\frac{1}{2}$$. AB.AC=^{}$ $\frac{1}{2}$$.29,7.19≈282,15^{}$ (đvdt)
Bài 5)
$cosA=^{}$ $\frac{AC}{AB}$=$\frac{5}{13}$
$sinA=√(1-(cosA)^2)=^{}$ $\frac{12}{13}$=$\frac{BC}{AB}$
$⇒ AC=^{}$ $\frac{5AB}{13}$, $BC=^{}$ $\frac{12AB}{13}$
$⇒ tanB=^{}$ $\frac{AC}{BC}$=$\frac{5}{12}$
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