Xét △ABC vuông tại A
\(\sin B=\dfrac{AC}{BC}\)
⇔\(\sin\dfrac{1}{2}=\dfrac{AC}{10}\)
⇒AC \(\approx\) 0.1 cm
Có: \(AB^2+AC^2=BC^2\) (định lí Pytago)
⇔AB2 +0.12=102
⇒AB=\(\sqrt{10^2-0.1^2}\) \(\approx10\)
❤\(\sin C=\dfrac{AB}{BC}=\dfrac{10}{10}=1\)
\(\cos C=\dfrac{AC}{BC}=\dfrac{0.1}{10}=0.01\)
\(\tan C=\dfrac{AB}{AC}=\dfrac{10}{0,1}=100\)
\(\cot C=\dfrac{AC}{AB}=\dfrac{0.1}{10}=0.01\)