Answer: In ΔLBN, LB ≈ 8.9 cm (8.925 cm), ∠B = 40°, BN = 8.4 cm.
Analysis: \(\dfrac{AB}{LB}=\dfrac{BC}{BN}=\dfrac{AC}{LN}=\dfrac47\), so ΔLBN is bigger. Vertex B is common, so L lies on ray BA and N on ray BC.
\(BN = 4.8\times\tfrac74 = 8.4\) cm, \(LB = 5.1\times\tfrac74 = 8.925\) cm.
Steps of construction:
- Draw seg BC = 4.8 cm. At B draw \(\angle CBA = 40^\circ\) and mark A with BA = 5.1 cm. Join AC. This is ΔABC.
- Draw ray BX making an acute angle with BC, on the side opposite to A. Mark 7 equal parts \(B_1,\dots,B_7\).
- Join \(B_4C\). Through \(B_7\) draw a line parallel to \(B_4C\), meeting ray BC (extended) at N.
- Through N draw a line parallel to CA, meeting ray BA (extended) at L.
- ΔLBN is the required triangle.