Construct a triangle abc in which base bc 5 cm ,angle b = 30° and ab +ac = 5.2​

Given, in ∆ABC, base BC = 3 cm, ∠B = 30° and AB+ AC = 5.2 cmSteps of construction(i) First, draw base BC = 3 cm.(ii) Construct∠YBC = 30° at the point B.(iii) From ray BY, cut-off line segment BD = AB+ AC = 5.2 cm.(iv) Join CD.(v) Draw the perpendicular bisector of CD, which meets BY at A(vi) Join AC. Thus, we get the required ∆ABC.Justification Since, A lies on the perpendicular bisector of CD.∴ AC = ADNow, BD = 5.2 cm =>BA+ AD = 52 cm∴ BA+ AC = 5.2 cm