What proportion of triangles are acute?

Question

An acute triangle is one in which all angles are acute, ie all angles $<90^\circ$. What proportion of triangles are acute?

I have two attempted answers which are different; I suspect this may be because the question is incoherent. I'd appreciate either an explanation of the correct answer or of why the question is insufficiently defined.

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Answer 1: $\frac{1}{2}$.

Since no more than one angle in a triangle can fail to be acute, whether the triangle is acute is determined by the value of one of its angles. Let $T$ be a triangle and pick an angle $\theta(^{\circ})$. Since $T$ is a triangle we must have $0<\theta<180$.

Now $T$ is acute iff $0<\theta<90$, so $\frac{1}{2}$ of all triangles are acute.

Answer 2: $\frac{1}{4}$.

(I lack the graphics skills to properly illustrate this.) Consider 3D Cartesian axes with $0< x,y,z< 180$. Now the points on the plane $x+y+z=180$ represents all possible triples $(\alpha,\beta,\gamma)$ that could be angles of a triangle. Note that this plane (or its closure I suppose, if we're being picky) has the shape of an equilateral triangle with vertices at $(180,0,0)$, $(0,180,0)$ and $(0,0,180)$.

Now we restrict the axes to $0<x,y,z<90$. This restriction of $x+y+z=180$ forms an equilateral triangle with vertices at $(90,90,0)$, $(0,90,90)$ and $(90,0,90)$ (the midpoints of the sides of the first triangle). But this new triangle's area is $\frac{1}{4}$ of the area of the previous triangle so $\frac{1}{4}$ of all triangles are acute.

Answer 1

Inspired by the article on Bertrand's paradox as provided by @Arthur, I offer this thought experiment:

Consider taking the isosceles right triangle and constructing a perpendicular bisector to its hypotenuse. Take the two endpoints of the right triangle and any point along the perpendicular bisector as a new triangle, as shown:

Any point chosen between the base and the height of the right triangle will result in an obtuse triangle, and any point chosen along the line beyond the height will make an acute triangle. From this we can obviously see, there are infinitely many more acute triangles than obtuse.

Answer 2

I'm going to offer this argument for the answer being $1/2$; Consider two points that lie on the unit circle, one is fixed and one moves around the circle clockwise. A "random" isosceles triangle can be constructed by connecting the fixed point, the mobile point, and the center of the circle. As the mobile point moves around the circle, the resultant triangle oscillates between acute and obtuse, the changeover occurs when the mobile point crosses an imaginary diameter of the circle perpendicular to the radius of the fixed point. This imaginary diameter bisects the area of the circle, therefore the triangle is acute half the time and obtuse the other half. Therefore half of all (isosceles) triangles must be acute.