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The Ambiguous Case

Date: 04/01/2003 at 13:38:38
From: Les
Subject: The Ambiguous Case

I do not understand how to use the ambiguous case to determine the 
number of triangles that can be constructed. Is there a simple way to  
answer the following question:

How many triangles can be constructed if, for example, a=4, A=30, and 
c=12?  Or a=9, b=12, and A=35?

I am confused about how to do this.
Thank you,

Date: 04/01/2003 at 15:12:48
From: Doctor Rick
Subject: Re: The Ambiguous Case

Hi, Les.

We can start by applying the law of sines. In your first example, we 

  sin(C)/c = sin(A)/a
  sin(C)/12 = sin(30)/4
  sin(C) = sin(30)*12/4
  sin(C) = 0.5*12/4
  sin(C) = 1.5

You know that the sine of any angle is between -1 and 1. There is no 
angle whose sine is 1.5, therefore there are no solutions in this 

Let's consider another example: B = 50 deg, b = 12, c = 10. In this 

  sin(C)/c = sin(B)/b
  sin(C)/10 = sin(50)/12
  sin(C) = sin(50)*10/12
  sin(C) = 0.638370
  C = 39.67 degrees

Is this the only solution? Not necessarily, because there is another 
angle whose sine is 0.638370, namely, 180 - 39.67 = 140.33 degrees. 
Can we have a triangle with angle B = 50 deg and angle C = 140.33 deg? 
No, because B+C = 190.33 deg, which is more than the sum of all three 
angles of any triangle (180 deg). Thus this case is not ambiguous: 
there is exactly one triangle that satisfies the conditions. The one 
solution has angle C = 39.67 degrees.

If the law of sines gives a sine of the missing angle that is less 
than 1, AND both angles with this sine (the arcsine of the angle and 
180 minus the arcsine) are less than 180 minus the known angle, then 
there are two triangles that satisfy the conditions.

Here are some more thoughts:

You can recognize whether an SSA specification of a triangle has 0, 1 
or 2 solutions without going through the law of sines. Consider the 
geometry of the triangle. If we have a base (b) of known length and a 
known angle A, the third vertex (B) must lie along a ray:

                 B  /
             D  /  \
              /     \
         B' /  \     \a
          /      \    \
        /   \      \   \
      /        a\    \  \
    /               \  \\
 A           b           C

We know the length of the side a, and we want to find a point (or 
points) B along the ray such that BC = a. 

Consider the perpendicular CD from C to the ray. Its length is 
b*sin(A), and it is the shortest distance from C to the ray. If 
a < b*sin(A), then we know that no point on the ray will be a distance 
a from C; all points are farther than this. Thus the condition for no 
solution is

  a < b*sin(A)

In the figure, I show the ambiguous case: both CB and CB' have length 
a. The triangle BCB' is isosceles, and CD is its altitude, so BD = 
B'D. This is the ambiguous case: both ABC and AB'C satisfy the 

If we increase length a, then point B moves out along the ray, but B'
moves in toward A. When a = b, point B' is coincident with point A. 
If a > b, we no longer have an ambiguous case: only triangle ABC 
satisfies the conditions, because B' is in effect pushed off the ray, 
onto the ray in the opposite direction.

Thus, the three cases can be distinguished easily:

  a < b*sin(A)       no solutions
  b*sin(A) < a < b   two solutions
  b < a              one solution

In my figure, A is an acute angle. What happens if A is obtuse? It's 
easy to see that there is no ambiguity. If a > b, there is one 
solution; if a < b, there are no solutions.

- Doctor Rick, The Math Forum 
Associated Topics:
High School Triangles and Other Polygons
High School Trigonometry

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