$\displaystyle{\cot{{Y}}}=\frac{{1}}{{{\tan{{Y}}}}}=\frac{{1}}{{\frac{{21}}{{20}}}}=\frac{{20}}{{21}}\usepackage{cmbright}\Huge\$

Now, from the definitions: $\displaystyle{\cos{{Y}}}=\frac{{{a}{d}{j}}}{{{h}{y}{p}}}=\frac{{20}}{{29}}\usepackage{cmbright}\Huge\$ and $\displaystyle{\tan{{Y}}}=\frac{{{o}{p}{p}}}{{{a}{d}{j}}}=\frac{{21}}{{20}}\usepackage{cmbright}\Huge\$

A triangle with $\displaystyle{a}{d}{j}={20}\usepackage{cmbright}\Huge\$ , $\displaystyle{h}{y}{p}={29}\usepackage{cmbright}\Huge\$ , and $\displaystyle{o}{p}{p}={21}\usepackage{cmbright}\Huge\$ will have these ratios.

So: $\displaystyle{\sin{{Y}}}=\frac{{{o}{p}{p}}}{{{h}{y}{p}}}=\frac{{21}}{{29}}\usepackage{cmbright}\Huge\$