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Why the method of images doesnt work with two semi infinte conductors with angle of 70º...

Why the method of images doesnt work with two semi infinte conductors with angle of 70º or 20º between them? I know the fictitious charge isnt suppose to be in the region you want to calculate the potential, but i can figure out the right configuration to these systems.

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The method of images fails in this case as, at the certain angles specified, you cannot substitute the surface by adding an additional point charge, which means creation of an image is not possible in this case, so we cannot calculate the resultant potential due to the two semi infinite conductors. Treating this system with respect to other orientations of the two conductors, will help it to be solved.

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