I was writing an answer to this question, which asks about what happens to the apex of an isosceles triangle if a vertex is at infinity. I thought it would be very easy to prove it by setting up a limit of a function of the hypotenuse as it goes to infinity, but it turns out to be 180 degrees and not 0 as one would expect. What's wrong with my reasoning?
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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, the height is the side adjacent to the angle $a/2$, and it'll be denoted by $d$. The relation between the sides and the angle is (by definition of cosine): $$\cos{(a/2)} = \frac{d}{h}$$ Using the double angle formula for cosine to find an expression in terms of $a$, we get:
$$\sqrt{\frac{1 + \cos{(a)}}{2}} = \frac{d}{h}$$ I chose the positive value only instead of $\pm$ since the sides are always positive. Now, we solve for $a$ to get a formula for the angle from the two sides:
$$a = \cos^{-1}{\left(\frac{2d^2}{h^2} - 1\right)}$$ From here, we can take the limit of the function as $h$ goes to infinity: $$\lim_{h \to \infty} \cos^{-1}{\left(\frac{2d^2}{h^2} - 1\right)}$$
The limit is equal to 180 degrees, not 0. If it were a $+1$ instead, the limit would be 0. Is there a glaring mistake I'm missing or is it something subtle?
Thank you for your time.
