xét tam giác ABC có \(BC^2=AB^2+AC^2=>BC=\sqrt{AB^2+AC^2}=\sqrt{8^2+15^2}=17\)
\(sinB=\frac{AC}{BC}=\frac{15}{17}=>cosC=\frac{15}{17}\)
\(cosB=\frac{AB}{BC}=\frac{8}{17}=>sinC=\frac{8}{17}\)
\(tanB=\frac{AC}{AB}=\frac{15}{8}=>cotgC=\frac{15}{8}\)
\(cotgB=\frac{AB}{AC}=\frac{8}{15}=>tanC=\frac{8}{15}\)