I'll try to answer this question in terms of definitions of the trigonometric functions in which we keep everything on the unit circle.
As an aside, I do feel compelled to point out that the question as currently edited is completely incompatible with a unit-circle construction of the functions, but based on your comments I am willing to believe that you did intend to use the unit circle and that you only used the word "integer" in your edited question by mistake.
In the unit-circle definitions of the trigonometric functions, if you do them correctly, the only reason to mention a "triangle" is to show how the unit circle definitions agree with the SOH-CAH-TOA triangle-based definitions
when the angle is between zero and a right angle.
If you had never seen the SOH-CAH-TOA definitions and instead had used only the unit-circle definitions, there would be no need for you ever to think about a triangle in relation to the definitions of these functions.
To use the unit circle, you construct an angle $\theta$ counterclockwise from the positive $x$ axis. The ray at that angle outward from the origin intersects the circumference of the circle at some point.
Let $(x,y)$ be the coordinates of that intersection point.
Then
\begin{align}
\cos\theta &= x, \\
\sin\theta &= y, \\
\tan\theta &= \frac yx. \\
\end{align}
Forget "opposite" and "adjacent"; all you need is $x$ and $y.$
Now suppose you want $\tan\theta = v,$ where $v$ is some arbitrary real number, possibly very large.
Let's see how you can find the exact values of $x$ and $y$ you need in order to construct an angle $\theta$ such that $\tan\theta = v.$
If $(x,y)$ is a point on the unit circle, we know that $x^2 + y^2 = 1.$
From the definition of the tangent function, we know that if the ray
at angle $\theta$ intersects the unit circle at $(x,y),$ then
$\tan\theta = \frac yx.$ That is, $v = \frac yx.$
Now observe that
$$
v^2 + 1 = \frac {y^2}{x^2} + 1 = \frac{y^2 + x^2}{x^2} = \frac 1{x^2}. \tag T
$$
In general, $x$ in this equation could be positive, negative, or zero, but let's just try non-negative values of $x$ for now.
Solving Equation $(T)$ for $x,$ taking only the non-negative solution
and discarding any negative solution, we get
$$
x = \frac 1{\sqrt{v^2 + 1}}.
$$
So that's $x.$ And of course since $v = \frac yx,$ and we now know both
$v$ and $x,$ we can solve for $y$:
$$
y = vx.
$$
That's all we need! Set $x$ and $y$ to these values (computed from the
given tangent value $v$), draw the ray from the origin through
$(x,y),$ and you have constructed the angle whose tangent is $v.$
The secret to getting a large tangent is that even though we have restricted both $x$ and $y$ to be in the range $-1$ to $1,$ so neither of them can get very large, there is nothing to stop us from making $x$ very small.
When $x$ is tiny, $y$ is near $1,$ and $\frac yx$ is a large number.