Incorrect. First, $\displaystyle{\sin{{B}}}=\frac{{3}}{\sqrt{{{58}}}}=\frac{{o}}{{h}}\usepackage{cmbright}\Huge\$ , so you could draw a right triangle with $\displaystyle{o}={3}\usepackage{cmbright}\Huge\$ and $\displaystyle{h}=\sqrt{{58}}\usepackage{cmbright}\Huge\$ . Next, $\displaystyle{\tan{{B}}}=\frac{{6}}{{14}}=\frac{{3}}{{7}}=\frac{{o}}{{a}}\usepackage{cmbright}\Huge\$ . In this triangle you would have $\displaystyle{a}={7}\usepackage{cmbright}\Huge\$ . Use the definitions or the reciprocal relationships to find the other trigonometic ratios.