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Law of Sines
Congruent triangles
Solving Triangles

Given two adjacent side lengths and an angle opposite one of them, there is no definite completion of a triangle. According to triangle congruence postulates, two triangles cannot be proved congruent given these three elements. This configuration is commonly referred to as side-side-angle (SSA).

With the SSA configuration, there is a fixed angle connecting the base of the triangle and one of the adjacent sides. The length of the base is unknown, denoted below by a dashed line. The length of the third side length is also fixed, but neither angle adjacent to that side is known. This means that this third side can swing from the upper vertex in any way that connects this vertex and any point along the indefinitely sized base so that all three sides of the triangle are connected.

Given any SSA configuration, there are several outcomes that can occur when solving a triangle: no solution, one solution, or two solutions. Which of the three scenarios occurs for any SSA configuration depends on the length of the swinging side compared to the height of the triangle.

In any SSA configuration, we can draw the height of the triangle even though we don't know the length of the base. The height is the perpendicular distance from the upper vertex to the base. Because there is a right angle between the height and base, we can always use the fixed angle and the length of the fixed side to determine the height. Below, we will see how we can determine the number of solutions by comparing the known length of the swinging side to this known height of the triangle.

Let $a$ be the length of the side opposite $A$ , an acute angle. The table below describes the different solutions for different scenarios.

Scenario	Number of Solutions	Type of Triangle
$a < h$	no solution	none
$a = h$	one solution	right triangle
$b > a > h$	two solutions	oblique
$a \geq b$	one solution	when $a = b$ , equilateral/isosocles when $a > b$ , obtuse

Because the SSA configuration can prompt different numbers of solutions for for different scenarios, it is often referred to as the ambiguous case.

We can use trigonometry to determine the value of height $h$ . Click below to see how.

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\sin A =\frac{opposite}{hypotenuse}

Substituting the appropriate variables,

\sin A = \frac{h}{b}

Multiply both sides by $b$ to get

h = b \sin A

First Scenario: No Solution

In the first scenario, the length of the swinging side is shorter than the height, $h$ . Because this side of the triangle is shorter than the height, there is no solution. The shortest distance between a point and a given line is the line segment that is perpendicular to the given line and goes through that point, which in this case is height $h$ . Since the side length is shorter than the shortest possible distance between the base and the upper vertex of the triangle, the side opposite the fixed angle will never be able to reach the base of the triangle. Thus, there are no solutions when the swinging side length $a$ is less than $h$ or $b \sin A$ .

In the picture below with numbers we've chosen, no matter how the orange side swings, it will never touch the base of the triangle. This triangle will never be complete.

h = b \sin A

Substituting in the appropriate measures,

h = 10 \sin 30^\circ

h = 10 (\frac{1}{2})

h = 5

Since the other given side length is $4$ and since $4<5$ , there is no solution.

In summary, when $a < b \sin A$ , there is no solution for a SSA configuration.

Second Scenario: One Right Solution

In the second scenario, the length of the swinging side is equal to the height, $h$ . Because this side of the triangle is equal to the height, there is only one solution. The height, as explained above, is the single shortest possible distance from the upper vertex to the base of the triangle. Since the swinging side of the triangle is the same length as the height, there is only one way to orient this leg to make the triangle complete: perpendicular to the base and through the upper vertex. Thus, there is only one solution when the length $a$ is equal to $h$ or $b \sin A$ , and this automatically forms a right triangle.

In the picture below, no matter how the green side swings, it will only touch the base of the triangle once. This triangle will only be complete when the triangle becomes a right triangle.

h = b \sin A

Substituting in the appropriate measures,

h = 10 \sin 30^\circ

h = 10 (\frac{1}{2})

h = 5

Since the other given side length is $5$ and since $5=5$ , there is only one solution which is a right triangle.

In summary, when $a = b \sin A$ , there is just one solution for a SSA configuration.

Third Scenario: Two Solutions

In the third scenario, the length of the swinging side is greater than the height, $h$ . Because this side of the triangle is greater than the height, there are two solutions to complete the triangle. The swinging side will complete the triangle at exactly two points: one in which the swinging side and the fixed side form an acute angle, and one where those same two sides form an obtuse angle. Thus, there are two solutions when the swinging side length $a$ is greater than $h$ or $b \sin A$

In the picture below, no matter how the blue side swings, it's endpoint will touch the base of the triangle only twice. This SSA configuration will complete two separate triangles.

h = b \sin A

Substituting in the appropriate measures,

h = 10 \sin 30^\circ

h = 10 (\frac{1}{2})

h = 5

Since the other given side length is $6$ and since $6>5$ , there is are two unique triangular completions.

In summary, when $a > b \sin A$ , there are two solutions for a SSA configuration.

Determining Both Solutions

The ambiguous case often produces two possible completions of the triangle. In these two potential triangles, the corresponding angles between the swinging sides and the unknown sides are supplementary. To find both triangles, use the law of sines to solve for the first triangle, then find the supplement of the measure of the angle between the swinging side and the base and solve using that angle.

Because of the cyclic nature of sine as a periodic function, the sine of a given angle is the same as the sine of its supplement. There are two solutions by the law of sines since

\frac{b}{\sin B} = \frac{b}{\sin (180^\circ -B)}

.

Fourth Scenario: One Oblique Solution

In this final scenario, the length of the swinging side is greater than the length of the fixed side. In this scenario, two possible triangles can be formed by swinging the side, but only one contains an angle of the given measure. To get one of the possible triangles, the lower endpoint of the swinging side swings past (to the left of) the vertex that connects the base and the fixed side so that the fixed angle is not included in the solution. This creates a triangle that includes the supplement of the fixed angle, but not the fixed angle. Because this fixed angle is missing from the completed triangle, this possibility is not a viable solution. Thus, there is only one solution when length $a$ is greater than length $b$ .

If $a=b$ , at least two of the sides of the triangle will be the same length and the solution will be either an isosceles or equilateral triangle.

In the picture below, the swinging pink side forms two triangles. The one on the left, however, does not include the fixed angle with a measure of 30 °, and is therefore not a solution for this SSA configuration.

In summary, when $a \geq b$ , there is one solutions for a SSA configuration.

Ambiguous Case Applet

Grab the point E to adjust the length of the swinging side. Grab the point B to move it around and change the height. Blue and purple triangles are solutions given the fixed angle A, the length of side AB, and the length of the swinging side. Red triangles are not viable solutions, like in the fourth scenario.

Pay attention to the numerical values on the side. You'll notice that the number of solutions depends on the length of BE compared to the height.

ERROR: Unable to find Java Applet file: AmbiguousCase.class.

Teaching Materials

Ambiguous Case Demonstration

During lessons about the ambiguous case, it can often be tricky to visualize the different solutions that a particular SSA configuration produces. Sometimes, it can be easy to forget that the length of the base is not fixed, which makes it easy to forget that the two different solutions can have two different-sized bases.

This activity will show the solutions for each of the scenarios when the swinging side length is changed. Students will be able to physically see the different possible solutions given an SSA configuration.

Materials Needed

Chalk/Chalkboard
Yarn
Tape
Scissors

Instructions

Draw a base of a triangle on the chalkboard, parallel to the ground. Indicate that it is not of a given length, perhaps by making a dotted or colored line all the way across the board.
Tape one end of the yarn at the end of the base. Continue to tape up the yarn so that the yarn follows what would be first fixed side of a triangle. Make sure that there is a big enough angle between these first two sides so that you can mark it. Now we have drawn the base of unknown length, the fixed angle, and the first fixed side, adjacent to the angles--the A and first S of our ASS triangle.
Tape the yarn to the board at some point above the dotted line. This will be the vertex between the two sides of fixed length. Let the leftover yarn hang straight down from the taped vertex. Make sure that the hanging yarn can swing back and forth. This will be the swinging side of fixed length. Be sure to leave enough yarn so that you can show all four scenarios.

Grab the end of the yarn and move it so that the swinging side rotates from the taped vertex. Wherever the end of the yarn touches the line, a triangle is completed.
First, show the scenario where the swinging side length is longer than the length of the other fixed side. Show how the fixed angle is not included in one of the completed triangles.
Now cut the yarn so that it is long enough to show the next case, when there are two unique solutions.
Cut the yarn again to show the scenario with one solution and a right triangle, and then again to show no solution.