Given the triangle below:
It can be observed that
[tex]m\angle ION=m\angle GOT(vertically\text{ opposite angles are equal)}[/tex][tex]\begin{gathered} \text{if,m}\angle GOT=168^0,then \\ m\angle ION=m\angle GOT=168^0 \\ m\angle ION=168^0 \end{gathered}[/tex]
It can also be observed that
[tex]\begin{gathered} m\angle AON+m\angle AOI=m\angle ION(given) \\ m\angle AOI=90^0(given,right-angle) \\ m\angle ION=168^0(derived\text{ earlier)} \\ So, \\ m\angle AON+90^0=168^0 \\ m\angle AON=168^0-90^0 \\ m\angle AON=78^0 \end{gathered}[/tex]
It can as well be observed that
[tex]m\angle NOG=180^0(angle\text{ on a straight line)}[/tex]
Hence,
m∠ION = 168°
m∠AON = 78°
m∠NOG = 180°
