GIẢI: 
sinB=3/4 =>cosC=3/5
Ta có: cos^2 C+sin^2 C=1 => sin^2C=1-(3/5)^2=7/16 
=>sinC=(√7)/5
=>tanC=sinC/cosC=[(√7)/5]/(3/)=(√7)/5

Có:

\(cosC=sinB=\dfrac{5}{13}\)

\(cosB=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\\ \Rightarrow sinC=\dfrac{12}{13}\)

\(tgC=\dfrac{sinC}{cosC}=\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}}=\dfrac{12}{5}\)

\(\Rightarrow cotgC=\dfrac{5}{12}\)

ΔCAB vuông tại C

=>\(CA^2+CB^2=AB^2\)

=>\(AB^2=1^2+2^2=5\)

=>\(AB=\sqrt5\)

Xét ΔCAB vuông tại C có

sin B=\(\frac{AC}{AB}=\frac{1}{\sqrt5}\)

cos B=\(\frac{BC}{BA}=\frac{2}{\sqrt5}\)

Xét \(\Delta ABC\) vg tại A có

BC\(^2\)= AC\(^2\)+AB\(^2\)( theo định lí Pi ta go)

\(\Rightarrow\)AB\(^2\)=BC\(^2\)-AC\(^2\)\(\Leftrightarrow\)AB\(^2\)=1,2\(^2\)-0,9\(^2\)=1,44 - 0,81= 0,63

\(\Rightarrow\)AB=\(\sqrt{0,63}\)=\(\dfrac{3\sqrt{7}}{10}\)

\(\Rightarrow\)sin B=\(\dfrac{AC}{BC}=\dfrac{0,9}{1,2}=\dfrac{3}{4}\)

    sinC=\(\dfrac{AB}{BC}=\)\(\dfrac{\dfrac{3\sqrt{7}}{10}}{1,2}\)=\(\dfrac{\sqrt{7}}{4}\)

mik nghĩ là sinC=0,8

                 CosC=0,6

                 tanC=\(\dfrac{\text{4}}{3}\)

                 cotgC=0,75

\(cosC=\dfrac{3}{5}\)

\(sinC=\dfrac{4}{5}\)

\(cotgC=\dfrac{3}{4}\)

\(tanC=\dfrac{4}{3}\)