At B assume member forces tensile. The 45° member AB has components $-F_{AB}/\sqrt2$ in both directions; BC acts downward. Horizontal equilibrium gives −10−F_AB/√2 = 0, so AB is compressive with magnitude 10√2. Its upward component is 10 kN; vertical equilibrium then requires BC to pull downward at 10 kN. Thus **10 kN tensile, B**.
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