Solution:

Let the segments have lengths $a \leq b \leq c \leq d \leq e$.

Suppose, for contradiction, that all triangles formed are obtuse-angled. Then, for any triangle with sides $x \leq y \leq z$, the triangle is obtuse only if $z^2 > x^2 + y^2$.

Consider the three largest segments: $c$, $d$, $e$.

- For triangle with sides $c$, $d$, $e$: $e^2 > c^2 + d^2$.
- For triangle with sides $b$, $c$, $d$: $d^2 > b^2 + c^2$.
- For triangle with sides $a$, $b$, $c$: $c^2 > a^2 + b^2$.

Adding these inequalities:

$e^2 + d^2 + c^2 > (c^2 + d^2) + (b^2 + c^2) + (a^2 + b^2)$

$e^2 + d^2 + c^2 > c^2 + d^2 + b^2 + c^2 + a^2 + b^2$

$e^2 + d^2 + c^2 > a^2 + 2b^2 + 2c^2 + d^2$

Subtract $d^2 + c^2$ from both sides:

$e^2 > a^2 + 2b^2 + c^2$

But $e \leq a + b$ (since any three segments form a triangle, so the sum of the two smallest is greater than the largest), so:

$e^2 \leq (a + b)^2 = a^2 + 2ab + b^2$

But $a^2 + 2ab + b^2 \leq a^2 + 3b^2$ (since $ab \leq b^2$), so:

$e^2 \leq a^2 + 3b^2$

But from above, $e^2 > a^2 + 2b^2 + c^2 > a^2 + 3b^2$ (since $c \geq b$), which is a contradiction.

Therefore, at least one of the triangles is acute-angled.