If ∠BOC is a right angle and $\mathrm{m}\angle \mathrm{COD}=58$, then $\mathbf{m}\angle \mathrm{DOE}=\underset{¯}{?}$, $\mathbf{m}\angle \mathrm{BOA}=\underset{¯}{?}$and data-custom-editor="chemistry" $\mathbf{m}\angle \mathrm{AOC}=\underset{¯}{?}$.

The given diagram is:

From the given diagram it can be noticed that $\angle BOC=\angle COE=90°$.

By using the angle addition postulate it can be obtained that:

$\angle COD+\angle DOE=\angle COE$.

Now,

$\angle COD+\angle DOE=\angle COE$

$58+\angle DOE=90$

$\angle DOE=90-58$

$\angle DOE=32$

$\mathrm{Therefore},\mathrm{the}\mathrm{measure}\mathrm{of}\mathrm{the}\mathrm{angle}\angle DOE\mathrm{is}32°.$

From the given diagram it can be noticed that the angles∠DOEand∠BOAare vertical angles.

Therefore, by using the vertical angle theorem it can be said that the angles∠DOEand∠BOAare the congruent angles.

Therefore, $\angle BOA=\angle DOE$.

Now,

$\angle BOA=\angle DOE=32°$

Therefore, the measure of the angle ∠BOA is 32°.

From the given diagram it can be noticed that the angles∠AOCand∠CODlies on the same straight-lineAD.

Therefore, the sum of the angles∠AOCand∠CODis 180°.

Therefore, $\angle AOC+\angle COD=180°$.

Now,

$\angle AOC+\angle COD=180$

$\angle AOC+58=180$

$\angle AOC=180-58$

$\angle AOC=122$

Therefore, the measure of the angle∠AOCis122°.