Question

Construct $∆$ABC such that $\angle$B = 70$°$, $\angle$C = 60$°$ and AB + BC + CA = 10.5 cm.

Answer 1

Steps of construction:
1. Draw DP = 10.5 cm.
2. Draw rays DY and PX, such that $\angle \mathrm{PDY}=35°\mathrm{and}\angle \mathrm{DPX}=30°$. The rays DY and PX intersect at A.
3. Draw the perpendicular bisector of seg DA, intersecting seg DP at point B.
4. Draw the perpendicular bisector of seg AP, intersecting seg DP at point C.
5. Join AB and AC.
7. Hence, $△$ABC is the required triangle.