Áp dụng định lý Py-ta-go trong tam giác ABC ta có:

AB = $\sqrt{AC^{2} + BC^{2}}$ = $\sqrt{0,9{2} + 1,2^{2}}$ = 1,5cm

Ta có: 

sinB = $\frac{AC}{AB}$ = $\frac{0,9}{1,5}$ = $\frac{3}{5}$

cosB =  $\frac{BC}{AB}$ = $\frac{1,2}{1,5}$ = $\frac{4}{5}$

tanB =  $\frac{AC}{BC}$ = $\frac{0,9}{1,2}$ = $\frac{3}{4}$

cotB =  $\frac{BC}{AC}$ = $\frac{1,2}{0,9}$ = $\frac{4}{3}$

Ta có: $\widehat{A}$ + $\widehat{B}$ = $90^{\circ}$ nên:

sin A = cosB =  $\frac{BC}{AB}$ = $\frac{1,2}{1,5}$ = $\frac{4}{5}$

cos A = sinB = $\frac{AC}{AB}$ = $\frac{0,9}{1,5}$ = $\frac{3}{5}$

tan A = cotB =  $\frac{BC}{AC}$ = $\frac{1,2}{0,9}$ = $\frac{4}{3}$

cot A = tanB =  $\frac{AC}{BC}$ = $\frac{0,9}{1,2}$ = $\frac{3}{4}$.