91Ó°ÊÓ

$\stackrel{\mathit{¯}}{\mathit{A}\mathit{H}}$in a given plane is considered.

Any point O between A and H is chosen.

Consider $\stackrel{\mathit{¯}}{\mathit{O}\mathit{A}}\mathit{\text{ }}\text{ a²Ô»å}\stackrel{\mathit{¯}}{\mathit{O}\mathit{H}}$and all these rays are drawn from O on one side of$\stackrel{\mathit{¯}}{\mathit{A}\mathit{H}}$.

These rays’ results in pairing with the real numbers from 0 to 180 in such a way that:

The line $\stackrel{\mathit{¯}}{\mathit{O}\mathit{A}}$is paired with 0, and $\stackrel{\mathit{¯}}{\mathit{O}\mathit{H}}$with 180.

If $\stackrel{\mathit{¯}}{\mathit{O}\mathit{B}}$is paired with $\mathit{x}$, and role="math" localid="1648996453337" $\stackrel{\mathit{¯}}{\mathit{O}\mathit{C}}$with $\mathit{y}$, then $\mathit{m}\mathit{∠}\mathit{B}\mathit{O}\mathit{C}\mathit{=}\left|x−y\right|$.

Thus, the figure is shown below, such that two lines are drawn through O that will form a$30°$ angle.

As it can be seen from the above figure, that on the plane of the paper through point O between A and B, there are maximum two lines that will make $30°$with $\stackrel{¯}{OB}$.

The line $\stackrel{¯}{OC}$can be drawn using protractor on one side of the $\stackrel{¯}{AB}$and $\stackrel{¯}{OD}$on the other side of the.

Therefore, only 2 lines can be drawn through point O to form$30°$ with $\stackrel{¯}{OB}$.