# Math Help - # of solutions for triangles

1. ## # of solutions for triangles

How do I check for the number of solutions in a triangle?

As in 1 solution, 2 solutions, or 0 solutions.

If Possible are there formulas to follow that can help me figure this out?

This here is the only stuff I have gotten to check the solutions, but what to do for other situations, I do not know:

If angle A is acute and a<b then there are 3 possibilities.
1. If a>b*sin(A) then there are 2 solutions
2. If a=b*sin(A) then there is 1 solution
3. If a<b*sin(A) then there are 0 solutions.

Wasn't sure how to make it work with obtuse triangles, or right triangles, or triangles that have different letters that are already identified.

2. ## Re: # of solutions for triangles

Welcome to MHF, Nickod777!

What kind of solutions do you mean?
Solutions for A?
Or solutions for a and b?

Presumably you mean something else, otherwise you would not get 0, 1, or 2 solutions...

3. ## Re: # of solutions for triangles

Originally Posted by Nickod777
How do I check for the number of solutions in a triangle?

As in 1 solution, 2 solutions, or 0 solutions.

If Possible are there formulas to follow that can help me figure this out?

This here is the only stuff I have gotten to check the solutions, but what to do for other situations, I do not know:

If angle A is acute and a<b then there are 3 possibilities.
1. If a>b*sin(A) then there are 2 solutions
2. If a=b*sin(A) then there is 1 solution
3. If a<b*sin(A) then there are 0 solutions.

Wasn't sure how to make it work with obtuse triangles, or right triangles, or triangles that have different letters that are already identified.
Just so that I'm sure I understand your question. To "solve" a triangle means to find all the unknown lengths and angles. Are you asking in what situations there could be 0, 1 or 2 possible solutions for a triangle?

4. ## Re: # of solutions for triangles

Like uh, I am trying to figure out this.

Like an easy way to check how many possible solutions there are for a triangle.

Like say for a 2 solution triangle, there are 2 possible solutions for each measurement that is missing.

Imagine you have a triangle. <B is 31*, side b is 8, side c is 13.

This is a SSA triangle.
I'm kinda lazy to write up all the information, so go here and look at example #1. It shows 1 possible case of answers, and another possible case of answer.
Solving SSA Triangles
This basically means that the triangle has 2 possible solutions, just because there's 2 solutions each for each missing measurement.

For triangles with 1 solution, it just means a triangle with its measurements having only 1 possible solution.

Go here, look at the triangle EX. 1 and it shows that the measurements that are missing only have 1 solution, no others. Solving ASA Triangles

And a triangle with no solutions, or 0 only means that the triangle cannot be solved with the Law of Cosines or Sines, or anything,
No Possible solutions to any of the measurements, or maybe 1 measurement could be off and wreck the whole triangle.

All I am asking for is basically a possibly easy way to check triangles quickly for how many solutions there are for them. 0, 1, 2.

Or say how to check how many solutions the triangle has quickly.

5. ## Re: # of solutions for triangles

As far as I've heard, I know of no way to quickly determine whether a missing piece of triangle information can have 0, 1, or 2 solutions. Besides, suppose we are certain that one missing element can only have one solution, but another might be one of two solutions. Would that then be a 1 or a 2 solution triangle?

I think a revised definition for "solutions" might involve the number of triangles that can meet the specified data. IOW, when we label a triangle as "2 solution", "1 solution", or "0 solution", we're stating that with the information given, can there be two triangles that meet the information, or only one, or none at all, respectively.

That said, one thing that may help you are the postulates and theorems from geometry with regard to proving congruence of triangles. That said, if the three pieces of information you have to work with are of the form:
Side Side Side (the lengths of all three sides - SSS)
Side Angle Side (the lengths of two sides and the angle between them - SAS)
Angle Side Angle (two angles and the length of the side in between them - ASA)
Angle Angle Side (two angles and the length of a side adjacent to one of them - AAS)
then those are "1 solution" triangles. Given a set of data meeting on of the forms above, there is only one triangle that will fit that data.

Any other combination of three pieces of triangle data:
Side Side Angle (the length of two sides and one angle adjacent to them - SSA)
will result in a "2 solution" triangle. There is more than one triangle that can fit that data.

Note, too, that if you have at least three of the six (three lengths of sides, or three angles) pieces of data for a triangle, you will never have a "0 solution" triangle. The only instance you will have zero solutions is if the data is not a valid triangle at all! For example: A triangle has sides of lengths 156, 87, and 68. This would be a "0 solution" traingle and you won't be able to find out what the angles are, because this "triangle" is really nothing of the sort: 87 and 68 add to only 155.

Hope this helps!

6. ## Re: # of solutions for triangles

Nice analysis!

I'd like to add that the 1-solutions always have exactly 1 solution, except the SSS case.
In the side-side-side case the sides need to fulfill the triangle inequalities: a+b>c, b+c>a, c+a>b.
Otherwise it has 0 solutions.

The 2-solutions case SSA can have 0, 1, or 2 solutions depending on the sides.
Suppose we have angle A and sides a and b.
Then the critical condition is that $\sin(A) \le {a \over b}$.
Equality meaning 1 solution.

7. ## Re: # of solutions for triangles

Thanks guys. I apologize if the question was a little unclear with the "solutions" and "triangles". Plus my book gave no specific way in solving them except for just figuring out all the sides+angles, which seemed like too long. So again, thanks guys.