(N/A) $105^{\circ} = 90^{\circ} + 15^{\circ}$
Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ With $O$ as the center and any suitable radius,draw an arc that meets $\overrightarrow{OA}$ at point $B$.
$III.$ With $B$ as the center and the same radius,draw an arc to cut the previous arc at point $C$. (This represents $60^{\circ}$).
$IV.$ With $C$ as the center and the same radius,draw another arc to cut the first arc at point $D$. (This represents $120^{\circ}$).
$V.$ Bisect the arc between $C$ and $D$ to get point $P$ such that $\angle AOP = 90^{\circ}$.
$VI.$ Bisect the arc between $P$ and $D$ to get point $Q$. The angle $\angle AOQ$ will be $105^{\circ}$ (since $90^{\circ} + 15^{\circ} = 105^{\circ}$).