Construct a triangle ABC, in which ∠B = 60°, ∠C = 45° and AB + BC + CA = 11 cm
Steps of Construction:
- 1.Draw a line segment PQ equal to AB + BC + CA = 11 cm.
- Make ∠ LPQ = ∠B = 60° and ∠ MQP= ∠C = 45°
- Bisect ∠ LPQ and ∠ MQP. Let these bisectors intersect at a point A .
- Draw perpendicular bisectors of line AP and AQ
- Let XY intersect PQ at B and RS intersect PQ at C.
- Join AB and AC
Thus, Δ ABC is a the required triangle