Bài 2:
a: \(\tan\widehat{B}=\dfrac{AC}{AB}\)
\(\Leftrightarrow AC=6\cdot\dfrac{5}{12}=2.5\left(cm\right)\)
b: \(BC=\sqrt{AB^2+AC^2}=\sqrt{2.5^2+6^2}=6.5\left(cm\right)\)
Bài 1:
a: \(BC=\sqrt{AB^2+AC^2}=\sqrt{1.6^2+3^2}=3.4\left(cm\right)\)
\(\sin\widehat{B}=\cos\widehat{C}=\dfrac{AC}{BC}=\dfrac{3}{3.4}=\dfrac{15}{17}\)
\(\cos\widehat{B}=\sin\widehat{C}=\dfrac{AB}{BC}=\dfrac{1.6}{3.4}=\dfrac{8}{17}\)
\(\tan\widehat{B}=\cot\widehat{C}=\dfrac{15}{8}\)
\(\cot\widehat{B}=\tan\widehat{C}=\dfrac{8}{15}\)
b: \(\sin^2\widehat{B}+\sin^2\widehat{C}\)
\(=\sin^2\widehat{B}+\cos^2\widehat{B}\)
=1
