Vì tam giác ABC vuông tại A nên:
AB2+AC2=BC2
Hay 32+42=BC2
=>BC=\(\sqrt{3^2+4^2}=5cm\)
a) \(sinB=\dfrac{AC}{BC}=\dfrac{4}{5}\)
\(cosC=\dfrac{AC}{BC}=\dfrac{4}{5}\)
\(tanB=\dfrac{AC}{AB}=\dfrac{4}{3}\)
b) cosB+sinC+2tanC= \(\dfrac{3}{5}+\dfrac{3}{5}+2.\dfrac{3}{4}=\dfrac{27}{10}\)