sinC bằng 1/2 góc nào bạn
Do \(\widehat{B}+\widehat{C}=90^0\Rightarrow sinC=cosB=\dfrac{1}{2}\)
Ta có: \(sin^2B+cos^2B=1\Rightarrow sin^2B=1-cos^2B=1-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}\)
Mà \(\widehat{B}< 90^0\Rightarrow sinB>0\Rightarrow sinB=\dfrac{\sqrt{3}}{2}\)
\(\Rightarrow tanB=\dfrac{sinB}{cosB}=\sqrt{3};cotB=\dfrac{1}{tanB}=\dfrac{1}{\sqrt{3}}\)