Construct an angle of $15^{\circ}$.
(N/A) Steps of construction:
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ Construct $\angle AOB = 60^{\circ}$.
$III.$ Draw $\overrightarrow{OC}$,the angle bisector of $\angle AOB$,such that $\angle AOC = \frac{1}{2} \angle AOB = \frac{1}{2}(60^{\circ}) = 30^{\circ}$.
$IV.$ Draw $\overrightarrow{OD}$,the angle bisector of $\angle AOC$,such that $\angle AOD = \frac{1}{2} \angle AOC = \frac{1}{2}(30^{\circ}) = 15^{\circ}$.
Thus,$\angle AOD = 15^{\circ}$ is the required angle.
$I.$ Draw a ray $\overrightarrow{OA}$.
$II.$ Construct $\angle AOB = 60^{\circ}$.
$III.$ Draw $\overrightarrow{OC}$,the angle bisector of $\angle AOB$,such that $\angle AOC = \frac{1}{2} \angle AOB = \frac{1}{2}(60^{\circ}) = 30^{\circ}$.
$IV.$ Draw $\overrightarrow{OD}$,the angle bisector of $\angle AOC$,such that $\angle AOD = \frac{1}{2} \angle AOC = \frac{1}{2}(30^{\circ}) = 15^{\circ}$.
Thus,$\angle AOD = 15^{\circ}$ is the required angle.
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