Draw $\Delta PQR$ with $m \angle P=60^{\circ}, m \angle Q=45^{\circ}$ and $PQ =6 \text{ cm}$. Then construct $\Delta PBC$ similar to $\Delta PQR$ whose sides have lengths $\frac{5}{3}$ times the lengths of the corresponding sides of $\Delta PQR$.

(N/A) Data: Construct $\Delta PQR$ with $m \angle P = 60^{\circ}, m \angle Q = 45^{\circ}$ and $PQ = 6 \text{ cm}$.
To construct: Construct $\Delta PBC$ similar to $\Delta PQR$ such that the ratio of their corresponding sides is $5:3$.
Steps of construction:
$(1)$ Construct $\Delta PQR$ with $m \angle P=60^{\circ}, m \angle Q=45^{\circ}$ and $PQ = 6 \text{ cm}$.
$(2)$ In the half-plane of $\overleftrightarrow{PQ}$ that does not contain $R$,draw a ray $\overrightarrow{PZ}$ such that $\angle QPZ$ is an acute angle.
$(3)$ With some appropriate radius and center $P$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_1$. With the same radius and center $P_1$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_2$ such that $P-P_1-P_2$. Similarly,with the same radius and center $P_k$,draw an arc to intersect $\overrightarrow{PZ}$ at $P_{k+1}$ such that $P_{k-1}-P_k-P_{k+1}$,where $k=2, 3, 4$.
$(4)$ Draw $\overline{P_3Q}$.
$(5)$ Through $P_5$,draw a ray parallel to $\overline{P_3Q}$ to intersect the extended ray $\overrightarrow{PQ}$ at $B$.
$(6)$ Through $B$,draw a ray parallel to $\overline{QR}$ to intersect the extended ray $\overrightarrow{PR}$ at $C$.
Thus,$\Delta PBC$ is the required triangle.
Justification: In $\Delta PP_5B$,$P-P_3-P_5$,$P-Q-B$ and $\overline{P_3Q} \parallel \overline{P_5B}$.
$\therefore \frac{PB}{PQ} = \frac{PP_5}{PP_3} = \frac{5}{3}$.
Similarly,in $\Delta PBC$,$P-Q-B$,$P-R-C$ and $\overline{QR} \parallel \overline{BC}$.
$\therefore \frac{PC}{PR} = \frac{PB}{PQ} = \frac{5}{3}$.