a: Xét ΔABC có \(AC^2=BA^2+BC^2\)
nên ΔBAC vuông tại B
b: \(\sin BAC=\dfrac{BC}{AC}=\dfrac{42}{58}=\dfrac{21}{29}\)
\(\cos BAC=\dfrac{40}{58}=\dfrac{20}{29}\)
\(\tan BAC=\dfrac{21}{20}\)
\(\cot BAC=\dfrac{20}{21}\)
c: \(BH=\dfrac{40\cdot42}{58}=\dfrac{840}{29}\left(cm\right)\)
\(BE=\dfrac{BH^2}{BA}=\left(\dfrac{840}{29}\right)^2:40=\dfrac{17640}{841}\left(cm\right)\)