Ambiguous Case Law of Sines Topic Index | Algebra2/Trig Index | Regents Exam Prep Center

In Geometry, we found that we could prove two triangles congruent using:

 SAS - Side, Angle, Side ASA - Angle, Side, Angle AAS - Angle, Angle, Side SSS - Side, Side, Side HL - Hypotenuse Leg for Right Triangles.
 We also discovered that SSA did not work to prove triangles congruent. We politely called it the Donkey Theorem  ; - )

By definition, the word ambiguous means open to two or more interpretations.
Such is the case for certain solutions when working with the Law of Sines.

 •  If you are given two angles and one side (ASA or AAS),    the Law of Sines will nicely provide you with ONE solution    for a missing side. •  Unfortunately, the Law of Sines has a problem dealing with SSA.    If you are given two sides and one angle (where you    must find an angle), the Law of Sines could possibly provide you    with one or more solutions, or even no solution.

Before we investigate this situation, there are a few facts we need to remember.

 Facts we need to remember: 1.  In a triangle, the sum of the interior angles is 180º. 2.  No triangles can have two obtuse angles. 3.  The sine function has a range of . 4.  If the = positive decimal < 1, the can lie in the first quadrant (acute <) or in the second quadrant (obtuse <).

Let's look at some cases.  In each example, decide whether the given information points to the existence of one triangle, two triangles or no triangles.

Example 1:  In , a = 20, c = 16, and m<A = 30º.  How many distinct triangles can be drawn given these measurements?

 Use the Law of Sines:

C = sin-1 (0.4) = 24º (to the nearest degree) - in Quadrant I.
Sine is also positive in Quadrant II.  If we use the reference angle 24º in Quadrant II,
the angle C is 156º.
But, with m<A = 30º and m<C = 156º the sum of the angles would exceed 180º.
Not possible!!!!

Therefore, m<C = 24º, m<A = 30º, and m<B = 126º and only ONE triangle is possible.

Example 2:  In , a = 7, c = 16, and m<A = 30º.  How many distinct triangles can be drawn given these measurements?

 Use the Law of Sines:

Since sin C must be < 1, no angle exists for angle C.
NO triangle exists for these measurements.

Example 3:  In , a = 10, b = 16, and m<A = 30º.  How many distinct triangles can be drawn given these measurements?

 Use the Law of Sines:

B = sin-1(.8) = 53.13010 = 53º.
Angles could be 30º, 53º, and 97º :  sum 180º
The angle from Quadrant II could create angles 30º, 127º, and 23º :  sum 180º

 TWO triangles possible.

This example is the Ambiguous Case.  The information given is the postulate SSA (or ASS, the Donkey Theorem), but the two triangles that were created are clearly not congruent.  We have two triangles with two sides and the non-included angle congruent, but the triangles are not congruent to each other.

 Topic Index | Algebra2/Trig Index | Regents Exam Prep Center Created by Frederick Roberts