Before a member can be cut, the reaction loads at the supports need to be determined. Since both members AB and CB are two force members, the direction of the reaction force at A and C are known. Considering the full structure, summation of forces at C gives

Member AB can now be cut member at a-a. At the cut, both the shear and moment internal forces are unknowns. The vertical and horizontal components of the force F_{AB} are

F_{AB-x} = F_{AB-y} = F_{AB} / cos45 = 150 N

Vertical and horizontal distances are shown on diagram. Summing moments at the cut gives,

ΣM_{cut} = 0
M + 150(10)(1-cos25) - 150(10)(sin25) = 0

M = 493.4 N

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