Question

Cho ΔABC.Kẻ AH ⊥ BC biết AB = 5cm,BH = 3 cm,BC = 10 cm.

a) Biết \(\widehat{C}\) = 30 độ.Tính \(\widehat{HAC}\) ?

b) Tính độ dài các cạnh AH,HC,AC ?

Answer 1

a) Xét \(\Delta ABC\) có :

\(\widehat{AHC}+\widehat{HAC}+\widehat{ACH}=180^{^{^O}}\) (định lí PITAGO)

=> \(\widehat{HAC}=180^{^O}-\left(\widehat{AHC+}\widehat{ACH}\right)\)

=> \(\widehat{HAC}=180^{^O}-\left(90^{^O}+30^{^O}\right)\)

=> \(\widehat{HAC}=180^{^O}-120^{^O}\)

=> \(\widehat{HAC}=60^{^O}\)

b) Xét \(\Delta ABH\perp H\) (\(AH\perp BC\)) có :

\(AH^2=AB^2-BH^2\) (định lí PITAGO)

=> \(AH^2=5^2-3^2=16\)

=> \(AH=\sqrt{16}=4\left(cm\right)\)

Ta có : H ∈ BC

=> \(BH+HC=BC\)

Hay : \(3+HC=10\)

=> \(HC=10-3=7\left(cm\right)\)

Xét \(\Delta AHC\perp H\) có:

\(AC^2=AH^2+HC^2\) (Định lí PITAGO)

=> \(AC^2=4^2+7^2=65\)

=> \(AC=\sqrt{65}\)