a) \(\widehat{C}=90^0-\widehat{B}=90^0-70^0=20^0\)
\(sinB=\dfrac{AC}{BC}\Rightarrow AC=10.sin70^0\approx9,4\left(cm\right)\)
\(AB^2+AC^2=BC^2\left(Pytago\right)\)
\(\Rightarrow AB=\sqrt{BC^2-AC^2}=\sqrt{10^2-9,4^2}\approx3,4\left(cm\right)\)
b) Áp dụng HTL:
\(AH.BC=AB.AC\Rightarrow AH=\dfrac{AB.AC}{BC}=\dfrac{3,4.9,4}{10}\approx3,2\left(cm\right)\)
c) Áp dụng HTL trong tam giác ABD vuông tại B có đường cao BH:
\(AB^2=AH.AD\Rightarrow AD=\dfrac{AB^2}{AH}=\dfrac{3,4^2}{3,2}\approx3,6\left(cm\right)\)
\(BD^2=DH.AD\Rightarrow BD=\sqrt{DH.AD}=\sqrt{\left(AD-AH\right).AD}\)
\(=\sqrt{\left(3,6-3,2\right).3,6}\approx1,2\left(cm\right)\)
