It is known that the formula for the perimeter $P$ of an isosceles triangle on a plane is $$P=2L+B$$, where $L$ is the length of the leg and $B$ is the length of the base. Now, let us study some ratios.
For an equilateral triangle (recall that $L=B$), the ratio $L/B=1$, and the ratio $P/L=3$.
For a 5-5-6 triangle, $L/B=5/6$ (about $0.8333$) and $P/L=3.2$
The base $B$ can be an arbitrarily small positive real number, so the ratio $L/B$ has no upper bound. Recalling the perimeter formula above, $P$ approaches $2L$ as $B$ approaches $0$, so $P/L$ approaches $2$. Thus, $2$ is an exclusive lower bound.
The central angle of an isosceles triangle is strictly less than half a circle. The length of $B$ approaches $2L$ as the central angle approaches a half-circle. As such, $L/B$ approaches $1$, so $1$ is the exclusive lower bound for $L/B$. $P$ approaches $4L$ as $B$ approaches $2L$, so $P/L$ approaches $4$, which is the exclusive upper bound of $P/L$.
We can see it is possible for an isosceles triangle to have $L$ and $B$ such that $P/L=L/B$. Is this ratio rational?
Set $$P/L=L/B$$.
Assume this is rational. If so, we can find coprime integers $P$ and $L$ to satisfy this equation, so that $P/L$ is a fraction in lowest terms
Recall that
$$2L+B=P$$
Subtracting $2L$ from both sides, we get
$$B=P-2L$$
Substituting into the equation above:
$$\frac P L=\frac{L}{P-2L}$$
$L<P$, so $P-2L<L$. But we already assumed that $P/L$ is in lowest terms. We have a contradiction, and this ratio is irrational.
So what is this ratio?
We set $L=1$
Then
$$P=\frac{1}{P-2}$$
Multiplying by $P-2$
$$P(P-2)=1=P^2-2P$$
Subtracting 1
$$P^2-2P-1=0$$
$$P=1+\sqrt{2}$$
this is about $2.4142$
(This also shows that $1+\sqrt {2}-1=\sqrt{2}$ is irrational)
So, with leg lengths $1$, this triangle has a perimeter of $1+\sqrt{2}$, and a base length of $\sqrt{2}-1$ (about $0.4142$. The base angle is about $65.5302$ degrees, and the vertex angle is about $43.9396$ degrees.