Question

Draw $∆$ABC in which BC = 5cm, AB + AC = 8cm and $\angle$C $-$ $\angle$B = 40$°$

Answer 1

Steps of construction:
1. Draw BC = 5 cm.
2. Draw ray CX, such that $\angle \mathrm{BCX}=90°+20°\left(\mathrm{i}.\mathrm{e}.90°+\frac{\angle \mathrm{C}-\angle \mathrm{B}}{2}\right)$.
3. With B as centre and radius 8 cm, draw an arc, intersecting ray CX at point D.
4. Join B and D.
5. Draw a ray CY, such that $\angle \mathrm{BCY}=20°\left(\mathrm{i}.\mathrm{e}.\frac{\angle \mathrm{C}-\angle \mathrm{B}}{2}\right)$.
6. Draw the perpendicular bisector of seg CE, intersecting seg BD at point A.
7. Join A and C.
8. Hence, $△$ABC is the required triangle.