Draw $\overline{ AB }$ of length $5.9\,cm$ and divide it in the ratio $5: 7$ from $A$.

(N/A) Data: $\overline{ AB }$ of length $5.9\,cm$ is given.
To construct: $\overline{ AB }$ is to be divided in the ratio $5: 7$ from $A$.
Steps of construction:
$(1)$ In different half-planes of $\overleftrightarrow{ AB }$ containing $\overline{ AB }$,draw $\overrightarrow{ AX }$ and $\overrightarrow{ BY }$ such that $\angle XAB$ and $\angle YBA$ are congruent acute angles.
$(2)$ With some appropriate radius and centre $A$,draw an arc to intersect $\overrightarrow{ AX }$ at $A_{1}$. Similarly,with centre $A_{1}$ and the same radius,draw an arc to intersect $\overrightarrow{ AX }$ at $A_{2}$ such that $A-A_{1}-A_{2}$. Similarly,draw an arc with the same radius and centre $A_{k}$ to intersect $\overrightarrow{ AX }$ at $A_{k+1}$ such that $A_{k-1}-A_{k}-A_{k+1}$,where $k=2, 3, 4$. Thus,we obtain five points $A_{1}, A_{2}, A_{3}, A_{4}$ and $A_{5}$ on $\overrightarrow{ AX }$ such that $AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}$.
$(3)$ Now,with the same radius and beginning with point $B$ as centre,obtain seven points $B_{1}, B_{2}, B_{3}, B_{4}, B_{5}, B_{6}$ and $B_{7}$ on $\overrightarrow{ BY }$ such that $BB_{1} = B_{1}B_{2} = \dots = B_{6}B_{7}$.
$(4)$ Draw $\overline{A_{5}B_{7}}$ to intersect $\overline{ AB }$ at $M$.
Thus,$M \in \overline{ AB }$ is the point which divides $\overline{ AB }$ in the ratio $5: 7$.
Justification: Here $\frac{AA_{5}}{BB_{7}} = \frac{5}{7}$. Since $\overrightarrow{AX} \parallel \overrightarrow{BY}$,the correspondence $AA_{5}M \leftrightarrow BB_{7}M$ between $\Delta AA_{5}M$ and $\Delta BB_{7}M$ is a similarity.
$\therefore \frac{AM}{BM} = \frac{AA_{5}}{BB_{7}} = \frac{5}{7}$.