a) Áp dụng Pytago ta có:
AB2 + AC2 = BC2
=> AC2 = BC2 - AB2 = 117
=> \(AC=\sqrt{117}\)
\(\sin C=\frac{AB}{BC}=\frac{18}{21}=\frac{6}{7}\)
=> \(\widehat{C}\approx59^0\)
=> \(\widehat{B}\approx31^0\)
b) Áp dụng Pytago ta có:
AB2 + AC2 = BC2
=> BC2 = 136
=> \(BC=\sqrt{136}\)
\(\tan C=\frac{AB}{AC}=\frac{10}{6}=\frac{5}{3}\)
=> \(\widehat{C}\approx59^0\)
=> \(\widehat{B}\approx31^0\)