Xét tam giác \(ABC\) có
\(AB^2+AC^2=6^2+\left(4,5\right)^2=56,25=\left(7,5\right)^2=BC^2\)
=> \(\Delta ABC\) vuông tại A
sin B = \(\dfrac{AC}{BC}=\dfrac{4,5}{7,5}\)
\(=>\widehat{B}=36^o52'\)
\(sinC=\dfrac{AB}{BC}=\dfrac{6}{7,5}\\=>\widehat{C}=53^o8'\)