Ta có: $\widehat{B}+\widehat{C}=90^o$
$⇒sinB=cosC$
$⇒cosC=\dfrac{4}{5}$
Mà $sin^2C+cos^2C=1$
$⇒sin^2C+\dfrac{16}{25}=1$
$⇒sin^2C=\dfrac{9}{25}$
$⇒sinC=\dfrac{3}{5}$ (do $sinC>0$)
$⇒tanC=\dfrac{sinC}{cosC}=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}$
$⇒cotC=\dfrac{1}{tanC}=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$