Question

Construct $∆$ PQR, such that PQ = 5.7 cm, $\angle$P = $\angle$Q = 50$°$

Answer 1

Steps of construction:
1. Draw a line segment PQ = 5.7 cm.
2. Draw $\angle \mathrm{QPY}=50°$.
3. Draw another angle $\angle \mathrm{PQX}=50°$.
4. The rays PY and QX intersect at point R.

Thus, $△$PQR is the required triangle.