What happens to the angles of an isosceles triangle if one vertex is at infinity? My son and I were trying to decide whether an isosceles triangle can ever have 90 degree base angles. I would argue that if the two equal length sides are both infinitely long, they must have 90 degree angles, because any angle less than 90 degrees would result in a side that is not infinitely long. My son on the other hand argued that in this case it is not a triangle because the angles of a triangle have to add up to 180 degrees. My husband added that if the angles are 90 degrees the lines would not be touching, even at infinity. 
Who's right?
 A: If you have a vertex $C$ of an isosceles triangle $\triangle ABC$ such that the angles $\angle A\cong\angle B=90º$ then you would have two perpendicular lines to a segment that pass through a single point out of the segment:
$$\overline{CA}\perp\overline{AB} \text{ and }\overline{CB}\perp\overline{AB}.$$ This violates euclidean geometry.
A: I agree with your son, but with a slightly different argument. I my mind it doesn't make sense to talk about a triangle with infinitely long sides. (In the projective plane you could define it, but I don't see the point).
A: I appreciate that this is an old question, but I had the same thought the other night and thought it interesting to plot the change in angle as height increases. The equation would be as follows (if my Excel equation is anything to go by), where $\theta$ is the top angle of the isosceles triangle, and the length of the base of the triangle $ = 2$ (so that the line bisecting the base produces two right angle triangles of base length $= 1$):
$$\theta = \arcsin \left( \dfrac 1 {\sqrt{(1+\text {height}^2)}} \right)$$
By this equation, theoretically, an infinitely tall isosceles triangle would have two base angles $= 90^\circ$ and $\theta = 0^\circ$, effectively making the sides of the triangle parallel. This is, of course, impossible - then again, so are a lot of things when infinity is involved.
Graph of change in top angle (theta) of an isosceles triangle with base length 2 as height increases from 1 to 100
A: It depends where these triangle are. If they’re on the surface of the earth (assuming the earth is a perfect sphere), the angles must add up to greater than 180°, and there are plenty of isosceles triangles with two right angles — no need for infinite sides. For example, take any two points on the equator together with the north pole as vertices.
If the triangles are formed from three points in the usual coordinate plane (which has no “point at infinity”), then the angles do have to add up to 180°, and the sides have to be finite in length.
A: Short Answer: it depends on the surface (the geometry) you're working on.
Long Answer:
Historically, $(2D)$ geometry and its laws have been made with the assumption that the objects lie on a plane. This means that the surface all objects under consideration are embedded in is flat, and we get nice, neat Euclidean geometry. This is the geometry most people are familiar with, as it closely resembles the geometry of our world on a small scale.
However, this geometry relies on 5 axioms which are mostly self-evident. The fifth one, called the parallel postulate, is equivalent to the property that the sum of angles in a triangle is 180 degrees. Whenever we're working in this geometry, this must be true, and it is satisfied in a way in a triangle with two perpendicular angles. To see why, look at this diagram:

In an isosceles triangle, the height from the apex is also the angle bisector, hence the angle $a$ is divided into two equal parts. We want to see the behavior of $a$ as the equal sides get longer and longer. In each right triangle formed, we have the equal side as the hypotenuse, so we'll call it $h$. Similarly, half the base is the side opposite the angle $a/2$, and it'll be denoted by $o$. The relation between the sides and the angle is (by definition of sine):
$$\sin{(a/2)} = \frac{o}{h}$$
Using the double angle formula for cosine to find an expression in terms of $a$, we get:
$$\sqrt{\frac{1 - \cos{(a)}}{2}} = \frac{o}{h}$$
I dropped the $\pm$ since the sides are always positive. Now, we solve for $a$ to get a formula for the angle from the two sides:
$$a = \cos^{-1}{\left(1 - \frac{2o^2}{h^2}\right)}$$
From here, we can take the limit (if your calculus is a bit rusty, it's a tool we use to describe the behavior of functions while the variable is around a point) of the function as $h$ goes to infinity:
$$\lim_{h \to \infty} \cos^{-1}{\left(1 - \frac{2o^2}{h^2}\right)} = \cos^{-1}{(1)} = 0$$
Thus, when the hypotenuse gets infinitely long, the angle tends to 0, and the base angles are 90 degrees since the sum must be 180 degrees.
But what if the sum didn't have to be 180? In fact, there are geometries called non-Euclidean geometries where this fifth axiom is broken and the objects are on another 3D object besides a plane. In the case of spherical geometry (which is what we use over large distances on Earth since our planet is roughly a sphere), there are triangles with two right angles and a non-zero third angle, for example. Look at this drawing:

(source: scientificamerican.com)
It has three right angles, and finite side lengths. So, to sum up, in Euclidean geometry, extending the apex to infinity results in a zero angle and two 90 angles, but this is a rather degenerate case. In non-Euclidean geometries such as spherical geometry, such triangles with two 90 degree bases are common, and you can have any non-zero angle as the apex angle's measure.
A: Well, my thought would be simply that if you're going to allow a triangle to have infinite long sides, then why not also allow it to have infinitesimal angles? The vertex of this triangle is an "angle" of $0^\circ$, and thus (assuming Euclidean space) the "triangle" still has 3 angles, whose total is $180^\circ = 90^\circ + 90^\circ + 0^\circ$.
This also agrees with the limiting value: set up an isosceles triangle with base $b$ and height $h$, then the vertex angle is $\alpha = 2\arctan(\frac{b}{2h})$, so
$$
\alpha = \lim_{h\to\infty} 2\arctan\left(\frac{b}{2h}\right) = \lim_{h\to\infty} 2\arctan(0) = 0^\circ.
$$
In a similar vein, the base angle of the triangle would be $\beta = \arctan\left(\frac{2h}{b}\right)$, and the limiting value is
$$
\beta = \lim_{h\to\infty} \arctan\left(\frac{2h}{b}\right) = \lim_{h\to\infty} \arctan(\infty) = 90^\circ.
$$
Of course, there are good reasons not to consider such an object a well-defined triangle in the first place. In most contexts, I would just call it a line segment with parallel rays emanating in the same direction from its endpoints.
With that said, I don't think I'd buy an argument that accepts polygons with infinite sides, but then says this object fails to be a triangle because it lacks a third angle at the vertex. You can't really open the door to infinite sides but then slam it shut it for infinitesimal angles. If an angle of $0^\circ$ is "not an angle", well then, by the same token a side of $\infty$ is likewise "not a side" in the first place.
To your husband's argument, I'm afraid I both agree and disagree - "at" infinity, the lines would both never meet, and meet at a $0^\circ$ angle. Mathematically, it depends on which direction you take the limit from. Intuitively, it is because "lines intersecting" (or not) is a discrete, finitary concept, and we don't have the luxury of assuming that logic that applies to the finite will also apply to the infinite.
Ultimately, it seems to me that the nature of infinity this family quibble (oh, how I wish this were the type of thing my family disagreed about!) is really about. The answer is that infinite quantities have strange and counterintuitive properties. When it comes to infinity, we don't have the luxury of assuming that the usual finitary logic we are accustomed to reasoning with will continue to be valid.



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*(EDIT: interesting; I see most of the other answers have to do with Euclid's parallel postulate, spherical geometry.
In spherical geometry, you can indeed have a finite isosceles triangle with two right angles at its base - take, for example, the triangle whose base runs along the Earth's equator, and whose sides are the prime meridian and the $90^\circ\text{W}$ longitude line, running from the equator up to the North Pole. This triangle is finite and in fact has three right angles!
On the other hand, in hyperbolic geometry, you can fill up an entire (hyperbolic) plane entirely with triangles whose angles are $(0^\circ, 0^\circ, 0^\circ)$:

However, from the title of the question, What happens to the angles of an isosceles triangle if one vertex is at infinity?, I interpreted the question to be more about the nature of infinity and limits, than about geometry, per se. So rather than go down that road I came at the question from a slightly different angle (hehe).
