How to tell if a triangle has two solutions?

[Tyrion101]

I've never been quite sure? Is it just a case of trial and error? Or just knowing the limits of sin, cos and tan?

 

[RUber]

I am not sure what you are solving for in this triangle.
If you look at basic geometry texts about triangle congruence, you will find that various combinations of side lengths and interior angles uniquely determine a triangle. If you have 3 peices of information you might have a unique triangle.

 

[Tyrion101]

I wasn't aware of the term but thanks!

 

[phinds]

I still can't figure out what it is that you are asking.

 

[Mark44]

I still can't figure out what it is that you are asking.

I think I do.

In a problem for which you're asked to "solve" a triangle, you are typically given three pieces of information. If you are given two sides and the included angle, then the length of the third side will be unique. There is just one way that the ends of the two sides can be connected, and this fixes the length of the unknown side.

OTOH, if you are given two sides and an angle that is not the included angle (not the angle formed by the two sides), then it is usually the case that there will be two solutions for the unknown side. This situation often shows up when you use the Law of Sines. One reason for this is that $\sin(\pi/2 + A) = \sin(\pi/2 - A)$. Both of these angles have the same sine.

The best way to deal with all of these situations is to draw a sketch that uses the given information.

 

[HallsofIvy]

This question is about the "ambiguous case" when you are given the lengths of two sides of a triangle and one angle, NOT the angle between the two sides. Constructively you can do this. Draw a line and mark it to the length of one of the given sides. Construct the given angle at one end of that side. At the other end, set compasses to the length of the second side and mark off an arc at that distance.

There are three possibilities:
1) The arc might not cross the second line at all- the radius is too short. There is NO such triangle.
2) The arc might be tangent to the second line. This is the case where there is exactly one triangle- and it is a right triangle.
3) The arc might cross the second line in two different points. This is the case where there are two such triangles.

Frankly, the simplest way to do determine whether there is no such triangle, one, or two is to actually try to solve the triangle. The standard way to solve this problem is to use the cosine law. The cosine law, an extension of the Pythagorean theorem says, that, in any triangle, with side lengths a, b, c and angle C opposite side c, then $c^2= a^2+ b^2- 2ab cos(C)$. In this situation, we are given the lengths a and c, and the angle, C. Finding the third side, b, is equivalent to solving the quadratic equation $b^2- (2a cos(C))b+ (a^2- c^2)= 0$.

We can solve that using the quadratic formula:
$$\frac{2a cos(C)\pm \sqrt{4a^2 cos^2(C)+ 4a^2- 4c^2}}{2}= a cos(C)\pm\sqrt{a^2 cos^2(C)+ a^2- c^2}$$.

Whether there is zero, one, or two solutions to that, depends on the discriminant $a^2 cos^2(C)+ a^2- c^2$. If that is positive there must be two solutions and so two such triangles.