The Ambiguous Case and Acute Triangles

In math class, a bonus homework question asked, “Given SSA, what are the necessary and sufficient conditions to prove that the triangle will be acute?”

In other words, create a test for SSA (side-side-angle) that a triangle will pass if it is acute.

My proposition: Given SSA, it must be an isosceles triangle (S1 = S2) with a given angle > 45.

Proof (by exhaustion):

Let S1 = Adjacent side (w.r.t to A) and S2 = Opposite side.

If adj*Sin(A) > opposite, obviously there will be no resulting triangle.

If adj*Sin(A) < opp, but opp > adj, there will only be one, obtuse triangle.

If adj*Sin(A) < opp, and opp < adj, there will be TWO resulting triangles; since both of them are not acute, we cannot assume that, given SSA in this situation, the triangle is acute.

This leaves us with adj*Sin(A) < opp and opp = adj (that is, an isosceles triangle).

Since we also assume A > 45 and that the triangle is isosceles, we know that the angle on the bottom of the opposite leg is equal, we know that 2A > 90, and thus the last angle is acute. Therefore, it’s an acute triangle.

But I’m getting reports that this wasn’t the answer. Can anyone find where I went wrong?

Is A the angle formed by the two S sides or is it any angle?

Hopefully I can say just the right things here to keep this from becoming a homework service. If you’ve already turned the homework in and what we say here won’t affect your grade, let me know and I can be more explicit.

My impression is that the problem you solved is ill-posed. You state that

But the problem asked for “necessary & sufficient” conditions, i.e. write down a set of conditions that (a) guarantee that the triangle is acute, i.e. are “necessary”, and (b) are such that all acute triangles satisfy these conditions, i.e. are “sufficient”. In the case I quoted above, you said that there exist acute triangles which don’t satisfy your condition. So your condition is necessary, but not sufficient.

The setup sailor alluded to, on the other hand, is perfectly well-defined; you don’t have the “two possible triangles” problem. So the problem as asked could refer to one of two setups, for one of which the problem is meaningless and for the other of which it is. It’s a fair bet that the problem is implicitly assuming the second case.

Ack, sorry. In the last paragraph, that should read

I have turned in my homework.

Man, that sucks that I got it wrong because I misinterpreted the problem. :frowning:

I interpreted “necessary and sufficient” to mean that it was just the right amount of conditions, (i.e. no more than necessary).

sailor, let’s see if I can represent it in ascii code:

                  / \  
            s1 /    \   s2 
                /      \
            < A

Of course, given just SSA, there can be two possible triangles.

Yeah, the “necessary and sufficient” thing is a little odd in this case. If you set up the problem in your case, it’s a little like asking, “Given the hair color of a person, find necessary and sufficient conditions that they are female.” Since the data that you’re being told to consider (a given hair color) could equally well correspond to a woman or a man, the problem isn’t well-posed. Replace “hair color of a person” with “SSA of a triangle” and “female” with “acute” and you’ve got your original problem back.

If your teacher is a nice guy/gal, you can probably argue for some partial credit: You did make a good effort to solve the problem, and the problem (if it was posed as you said it was in the OP) is a little ambiguous. OTOH, it is only a bonus problem, and so is probably not worth raising a huge stink over.

Out of curiosity, has your teacher distinguished between SSA (side, then side, then angle) and SAS (side, then angle, then side) in class? If so, then your case is a little stronger.

Since you’ve already turned in your homework, I’ll spell out the answer.

[symbol]If you have two sides and the included angle, you can use the law of cosines to solve for the third side, and then you can use the law of cosines again to solve for the other two angles. You will need to use the fact that the sum of the angles is constant.[/symbol]

ultrafilter’s method would probably work, but there’s a simpler way to do it than that. Draw the segment XY, with length S[sub]1[/sub]. Then draw a ray passing through the point X at an angle A to the segment XY.

Now ask yourself: where on this ray must the third point Z be to make this triangle acute? To answer this draw a perpendicular from the ray passing through Y; call the intersection of this segment and the ray W[sub]1[/sub]. Also draw a perpendicular to the segment XY passing through Y; call the intersection of this perp. and the ray W[sub]2[/sub]. It’s not too hard to see that if Z is closer to X than W[sub]1[/sub], then XZY is obtuse; and if Z is farther than W[sub]2[/sub] from X, then XYZ is obtuse. This gives you a possible range of values for the second side, and the distances XW[sub]1[/sub] and XW[sub]2[/sub] can be found via simple trig.

Of course, it looks ten times more complicated when put into words. Try drawing out the figure and you’ll see what I mean.

Massive hint:
Use SSA construct a test for a right triangle, That’s the boundary condition between an acute and an obtuse triangle.

I must say that the way it was worded I am not at all sure what is being asked here. Here is how the construction goes: Given SSA, first construct the angle A and then mark off the distance, the second S, along one of the sides to get a second vertex B. Open a compass to a distance of the first S and place it at B. Now draw a circle of that radius. There are three possibilities.

  1. It might not intersect the third side at all. There is no triangle.

  2. It might intersect the third side twice. Then there will be two triangles, one with an acute angle at B and one with an obtuse angle.

  3. Or it might be tangent to the third side and then there will be a unique solution and it will be a right triangle.

So which? I have to leave something to you to do. Here is the answer:

Let the given angle be A, the adjacent side have length b and the opposite side length a. Then using the law of cosines, you get a quadratic equation for the third side whose discrimanant is sqrt(-4b^2 cos^2 A + 4a^2). After some manipulation the needed condition emerges: if a/b < sin A, then we have case 1, if a/b < sin A, we have case 2 and if a/b = sin A, then we are in case 3.