Question

$∆$ABC $~$ $∆$LMN. In $∆$ABC, AB = 5.1 cm, $\angle$B = 55$°$; $\angle$C = 65$°$ and $\frac{AC}{LN}=\frac{3}{5}$; then construct $△$ LMN.

Answer 1

$$\begin{gathered} \angle \mathrm{ABC}+\angle \mathrm{ACB}+\angle \mathrm{BAC}=180° \\ \Rightarrow \angle \mathrm{BAC}=180°-\angle \mathrm{ACB}-\angle \mathrm{ABC} \\ \Rightarrow \angle \mathrm{BAC}=180°-65°-55° \\ \Rightarrow \angle \mathrm{BAC}=60° \end{gathered}$$

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}∆\mathrm{ABC}\mathrm{is}\mathrm{similar}\mathrm{to}∆\mathrm{LMN}. \\ \therefore \frac{\mathrm{AB}}{\mathrm{LM}}=\frac{\mathrm{AC}}{\mathrm{LN}}=\frac{\mathrm{BC}}{\mathrm{MN}} \\ \mathrm{Also},\frac{\mathrm{AC}}{\mathrm{LN}}=\frac{3}{5} \\ \therefore \frac{\mathrm{BC}}{\mathrm{MN}}=\frac{\mathrm{AB}}{\mathrm{LM}}=\frac{\mathrm{AC}}{\mathrm{LN}}=\frac{3}{5} \\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{LM}}=\frac{3}{5} \\ \Rightarrow \mathrm{LM}=\frac{5}{3}\times 5.1 \\ =8.5\mathrm{cm} \end{gathered}$$

Steps of construction:
1. Draw a line segment AB = 5.1 cm.
2. Draw $\angle$BAX=60$°$.
3. Draw another angle $\angle$ABY = 55$°$. The rays AX and BY intersect at point C.
4. Similarly, draw a line segment LM = 8.5 cm.
5. Draw $\angle$LMQ = 55$°$.
6. Draw another angle $\angle$MLP = 60$°$. The rays LP and MQ intersect at point N.

Thus, $∆$LMN is the required triangle, which is similar to $△$ABC.