a- consider an isosceles triangle $ABC$ with $AB=BC$, and $BH$ the altitude and $\frac{AC}{BH}\ge n>4$ and $n$ triangles equals to $ABC$ placed with each $BH_i$ segment parallel to the $AC$ segment and there is no way one can cover the $AC$ segments with those $n$ triangles.

b- Split up the triangle we wish to cover into $4$ smaller equilateral triangles with half the original side length. Then put one of the four movable triangles such that it shares its center with a unique one of the smaller equilateral triangles; as an equilateral triangle with side $\frac{s}{2}$ has the same circumradius as the inradius of an equilateral triangle with side length $s$, the smaller triangle will always fit inside the larger one.