a.
Ta có: \(\dfrac{AB}{AC}=\dfrac{3}{4}\Rightarrow AB=\dfrac{3}{4}AC\)
Áp dụng hệ thức lượng:
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Rightarrow\dfrac{1}{4,8^2}=\dfrac{1}{\left(\dfrac{3}{4}AC\right)^2}+\dfrac{1}{AC^2}\)
\(\Rightarrow\dfrac{1}{4,8^2}=\dfrac{1}{AC^2}\left(\dfrac{16}{9}+1\right)=\dfrac{25}{9}.\dfrac{1}{AC^2}\)
\(\Rightarrow AC^2=4,8^2.\dfrac{25}{9}=64\)
\(\Rightarrow AC=8\left(cm\right)\)
\(\Rightarrow AB=\dfrac{3}{4}AC=\dfrac{3}{4}.8=6\left(cm\right)\)
Áp dụng đl Pitago: \(BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)
\(sinB=\dfrac{AC}{BC}=\dfrac{3}{5}\Rightarrow B\approx36^052'\)
\(C=90^0-B=53^08'\)
b.
Áp dụng hệ thức lượng:
\(AB^2=BH.BC\Rightarrow BH=\dfrac{AB^2}{BC}=\dfrac{6^2}{10}=3,6\left(cm\right)\)
\(\Rightarrow S_{\Delta ABH}=\dfrac{1}{2}AH.BH=\dfrac{1}{2}.4,8.3,6=8,64\left(cm^2\right)\)
