GIVEN TWO SIDES AND A NON-INCLUDED ANGLE, HOW MANY TRIANGLES?

The ‘SSA’ triangle condition (two sides and a non-included angle), does not uniquely identify a triangle.
Given two positive real numbers (two side lengths) and a degree measure strictly between $\,0^\circ\,$ and $\,180^\circ\,$ (the angle),
there may be no triangle, exactly one triangle, or two triangles that match the SSA information.

When Will You Have Zero, One, or Two Triangles in an ‘SSA’ Situation?

Let $\,a\,$ and $\,b\,$ be positive real numbers, used to denote lengths of two sides of a triangle.
Let $\,\theta\,$ denote the measure of a non-included angle.
As shown at right, $\,a\,$, $\,b\,$ and $\,\theta\,$ form an ‘SSA’ situation—two sides and a non-included angle.

To significantly cut down on words, I say things like:
‘$\,a\,$ attaches to $\,b\,$’
instead of the more correct (but much wordier):
‘the side of length $\,a\,$ attaches to the side of length $\,b\,$’
Note that:
  • ‘SSA’ corresponds to ‘$\,a\,b\,\theta\,$’
  • $\,b\,$ attaches to $\,a\,$ at one end and has $\,\theta\,$ at the other end
  • $\,a\,$ is opposite $\,\theta\,$
To sketch an SSA situation for given values of $\,a\,$, $\,b\,$ and $\,\theta\,$:
  1. start by drawing a dashed line that will ‘hold’ the third (unknown) side of the triangle
  2. choose a point on the dashed line as the vertex for $\,\theta\,$
  3. attach the side of length $\,\color{green}{b}\,$ to the chosen vertex at angle $\,\color{orange}{\theta}\,$
  4. at the other end of $\,b\,$, attach a circle of radius $\,\color{purple}{a}\,$;
    the angle at which $\,b\,$ attaches to $\,a\,$ is initially unknown
The altitude shown (which has length $\,b\sin\theta\,$) is an important boundary between different behaviors—keep reading!




In the following progression of sketches, $\,b\,$ and $\,\theta\,$ are held constant, and $\,\theta\,$ is an acute angle.
The side of length $\,a\,$ gets bigger and bigger:

$0 < a < b\sin\theta$

When $\,a\,$ is less than $\,b\sin\theta\,$, it is too short to reach the dashed line.
No matter what the angle is between $\,a\,$ and $\,b\,$,
the triangle cannot be completed.

For $\,0 < a < b\sin\theta\,$, there is no triangle
determined by sides $\,a\,$ and $\,b\,$ and a non-included angle $\,\theta\,$.


$0 < a < b\sin\theta$

no triangle determined by $\,a\,$, $\,b\,$ and $\,\theta\,$

$a = b\sin\theta$

When $\,a\,$ reaches the altitude length ($\,a = b\sin\theta\,$)
then it is just long enough to hit the dashed line and form a (right) triangle.

For $\,a = b\sin\theta\,$, there is exactly one (right) triangle
determined by sides $\,a\,$ and $\,b\,$ and a non-included angle $\,\theta\,$.


$a = b\sin\theta$

exactly one (right) triangle determined by $\,a\,$, $\,b\,$ and $\,\theta\,$

$\,b\sin\theta < a < b\,$

This is the most interesting case!
This case is the reason there is no ‘SSA’ congruence theorem.

For values of $\,a\,$ strictly between $\,b\sin\theta\,$ and $\,b\,$,
the circle of radius $\,a\,$ intersects the dashed line at two different points.

Therefore, there are two different triangles that meet the SSA information.

For $\,b\sin\theta < a < b\,$, there are two different triangles
determined by sides $\,a\,$ and $\,b\,$ and a non-included angle $\,\theta\,$.

the two triangles
determined by
$\,a\,$, $\,b\,$ and $\,\theta\,$

$\,b\sin\theta < a < b\,$

two different triangles determined by $\,a\,$, $\,b\,$ and $\,\theta\,$

$a = b$

For $\,a = b\,$, there is exactly one isosceles triangle
determined by sides $\,a\,$ and $\,b\,$ and a non-included angle $\,\theta\,$.


$a = b$

exactly one isosceles triangle

$a > b$

When $\,a\,$ is greater than $\,b\,$, then the circle of radius $\,a\,$ intersects the dashed line
at two different points. However, only one of these points gives a triangle
with interior angle $\,\theta\,$.

For $\,a > b\,$, there is exactly one triangle
determined by sides $\,a\,$ and $\,b\,$ and a non-included angle $\,\theta\,$.


$a > b$

exactly one triangle

Summary: Boundaries for Zero, One, Two Triangles in an SSA Configuration

This graphic summarizes the ‘SSA’ situations discussed above.
Side $\,b\,$ and angle $\,\theta < 90^\circ\,$ are fixed.
The number line shows the values of $\,a\,$ corresponding to no, one, and two triangles.

The red (no $\,\triangle\,$), blue (two $\,\triangle\,$s), and black (one $\,\triangle\,$) intervals change
depending upon the values of $\,b\,$ and $\,\theta\,$.

Put in your own values for $\,b\,$ and $\,\theta\,$ below—have fun!

Type in desired values for $\,b\,$ and $\,\theta\,$ below.
Then, click the ‘Create Number Line Summary’ and ‘See the Triangle’ buttons.

  length of side $\,b\,$
  angle $\,\theta\,$ (in degrees, $\,0^\circ < \theta < 90^\circ\,$)


Note: If $\,\theta\,$ is an obtuse angle (or a right angle), then the situation is different.
In this case, no triangle exists until $\,a > b\,$, at which point there is a unique triangle:

Practice Sketching SSA Cases

Get a random SSA situation by clicking below.
Sketch the given situation on a piece of paper—make it roughly to scale, but don't actually measure anything.
Based on your sketch, make a conjecture as to how many triangle(s) meet this SSA configuration.
Then, check your conjecture.

Introduction to the Arcsine Function and Examples

You might be lamenting:

‘How am I going to remember all these different situations?’

Here's the good news:
The mathematics will tell you which situation you're in
(zero, one, or two triangles),
providing you do the mathematics correctly!

Let's illustrate with three similar examples.
In each, you're given an angle and its opposite side, plus a second side.
The key is this—recognize that it is an SSA situation, and be careful!
In all these examples, $\,\beta\,$ denotes the (initially unknown) angle across from the second side.

How many triangles have a $\,50^\circ\,$ angle with opposite side $\,4\,$, and another side of
length $\,7$?
  • Use the Law of Sines: $$ \begin{gather} \frac{\sin 50^\circ}{4} = \frac{\sin\beta}{7}\cr\cr\cr 4\sin\beta = 7\sin 50^\circ\cr\cr \sin\beta = \frac 74\sin 50^\circ \approx 1.34 \end{gather} $$
  • The sine function only takes on values in $\,[-1,1]\,$. Therefore, there is no angle $\,\beta\,$ with $\,\sin\beta = 1.34\,$.
  • No triangle exists that meets this SSA situation.
  • Using the notation of this section, we have $\,a = 4\,$, $\,\theta = 50^\circ\,$, and $\,b = 7\,$.

    We could have found $$\,b\sin\theta = 7\sin 50^\circ \approx 5.362\,$$ and noted that $\,a < b\sin\theta\,$, putting us in the ‘no triangle’ category.

    We didn't need to do this—the mathematics told us!
How many triangles have a $\,50^\circ\,$ angle with opposite side $\,6\,$, and another side of length $\,7$?
  • Use the Law of Sines: $$ \begin{gather} \frac{\sin 50^\circ}{6} = \frac{\sin\beta}{7}\cr\cr\cr \sin\beta = \frac 76\sin 50^\circ \approx 0.89 \end{gather} $$
  • There are two angles between $\,0^\circ\,$ and $\,180^\circ\,$ with sine equal to $\,0.89\,$, as shown:

  • We need to know:
    ‘What angle(s) have sine equal to $\,0.89\,$?’
    The arcsine function (abbreviated as ‘$\,\arcsin\,$’ but still pronounced ‘arc-sign’) helps us to ‘undo’ the sine function! It is discussed in detail in a future section—just briefly introduced here.
  • By definition, $$ \begin{align} \arcsin(x) &= \text{the angle between $\,-90^\circ\,$ and $\,90^\circ\,$}\cr &\qquad\qquad \text{whose sine is $\,x\,$} \end{align} $$
  • Therefore, $\,\arcsin(0.89)\,$ is the angle in the first quadrant whose sine value is $\,0.89\,$, as shown. To find the green angle with sine $\,0.89\,$, subtract the degree measure of $\,\arcsin(0.89)\,$ from $\,180^\circ\,$.
  • The possible triangle angle configurations are:
    $50^\circ$ $\beta = 63^\circ$ $180^\circ-50^\circ-63^\circ = 67^\circ$
    $50^\circ$ $\beta = 117^\circ$ $180^\circ-50^\circ-117^\circ = 13^\circ$
  • Using the notation of this section, we have $\,a = 6\,$, $\,\theta = 50^\circ\,$, and $\,b = 7\,$.
    Thus, $\,b\sin\theta = 7\sin 50^\circ \approx 5.362\,$.

    We could have noted that: $$\overbrace{\strut b\sin\theta}^{5.362} < \overbrace{\strut a}^{6} < \overbrace{\strut b}^{7}\,$$ This puts us in the ‘two triangle’ category. We didn't need to do this—the mathematics told us!
How many triangles have a $\,50^\circ\,$ angle with opposite side $\,8\,$, and another side of length $\,7$?
  • Use the Law of Sines: $$ \begin{gather} \frac{\sin 50^\circ}{8} = \frac{\sin\beta}{7}\cr\cr\cr \sin\beta = \frac 78\sin 50^\circ \approx 0.67 \end{gather} $$
  • There are two angles between $\,0^\circ\,$ and $\,180^\circ\,$ with sine equal to $\,0.67\,$, as shown:

  • The angle $\,\arcsin(0.67)\approx 42^\circ\,$ is the angle in the first quadrant whose sine value is $\,0.67\,$, as shown.
  • The other angle between $\,0^\circ\,$ and $\,180^\circ\,$ with sine equal to $\,0.67\,$ is $\,180^\circ - 42^\circ = 138^\circ\,$.
  • The potential triangle angle configurations are:
    $50^\circ$ $\beta = 42^\circ$ $180^\circ-50^\circ-42^\circ = 88^\circ$
    $50^\circ$ $\beta = 138^\circ$ $180^\circ-50^\circ-138^\circ = -8^\circ$
  • The second configuration clearly doesn't work!
    There is only one triangle that satisfies this SSA information.
  • Using the notation of this section, we have $\,a = 8\,$, $\,\theta = 50^\circ\,$, and $\,b = 7\,$.

    We could have noted that $\,a > b\,$, putting us in the ‘one triangle’ category. We didn't need to do this—the mathematics told us!

Comments about Notation When Solving Triangles

When naming triangles, it is common to use:
  • a capital letter (like $\,D\,$) to denote a vertex
  • a lowercase letter (like $\,d\,$) to denote the length of the opposite side
The vertex uniquely identifies the angle at that vertex, and we use this fact to simplify notation.
When we write things like ‘$\,\sin D\,$’ or ‘$\,180^\circ - D\,$’, then $\,D\,$ refers to the measure of the angle at vertex $\,D\,$.

Precalculus triangle notation is simpler than Geometry triangle notation.
Due to the nature of the work in Geometry,
careful distinctions need to be made between points, angles, and numbers:
  • A vertex is a point.
    A vertex is denoted by a capital letter, like $\,P\,$.
  • An angle is a geometric object—the union of two rays with a common endpoint.
    If there's only one angle at a vertex $\,P\,$, then the angle can be denoted using the symbol $\,\angle P\,$.
  • The measure of an angle is a real number.
    The measure of $\,\angle P\,$ can be denoted by $\,m\angle P\,$.
In Precalculus these distinctions are usually not necessary.
It is a nuisance to have to write things like ‘$\,\sin(m\angle P)\,$’.
Cumbersome notation (without a need for such) can make simple things look hard.

So—in Precalculus—when there is no possible confusion,
we use simple notation and let context determine the correct interpretation.
Here are some Precalculus examples:
  • In the expression ‘$\cos D\,$’, $\,D\,$ refers to the measure of an angle.
  • In the expression ‘$\,D = 180^\circ - E - F\,$’, $\,D\,$ (and $\,E\,$ and $\,F\,$) refer to measures of angles.
  • In the expression ‘$\,\triangle DEF\,$’, $\,D\,$ (and $\,E\,$ and $\,F\,$) refer to points (vertices).

A Complete Example:
Using the Law of Sines to Investigate an ‘SSA’ Configuration

Suppose a triangle has a $\,27^\circ\,$ angle with opposite side of length $\,23\,$.
Another side has length $\,30\,$. If possible, solve the triangle.
Round angles to one decimal place and sides to two decimal places.
Any value used inside an arcsine should be rounded to at least four decimal places.

SOLUTION:

It's useful to make a sketch to define notation.
(Try to do it approximately to scale, but certainly don't measure anything!)
Note that this is an ‘SSA’ situation, and be careful!
Make a conjecture about the number of triangles determined by the SSA information.

Looks like a two triangle situation!
It's also useful to use a table (or tables) to organize results:

ANGLEOPPOSITE SIDE
$\, D = 27^\circ$$d = 23$
$\, E = \,?$$\,e = \,?$
$\, F = \,?$$\,f = 30$

The table(s) will be filled in as computations are completed.

In this example, the tables are repeated,
so you can easily see each step where information is added.

When doing problems like this on your own,
you won't repeat the table(s)—you'll just fill them in as you go.

  • Use the Law of Sines: $$ \begin{gather} \frac{\sin 27^\circ}{23} = \frac{\sin F}{30}\cr\cr\cr \sin F = \frac{30}{23}\sin 27^\circ \approx 0.5922 \end{gather} $$
  • Find both angles (between $\,0^\circ\,$ and $\,180^\circ\,$) that have a sine of $\,0.5922\,$: $$ \begin{gather} \arcsin(.5922) \approx \color{red}{36.3^\circ}\cr\cr 180^\circ - 36.3^\circ = \color{red}{143.7^\circ} \end{gather} $$
  • Note:
    • If your calculator is in radian mode, then $\,\arcsin(0.5922)\,$ is reported in radians.
    • If your calculator is in degree mode, then $\,\arcsin(0.5922)\,$ is reported in degrees.
    • WolframAlpha reports both radians and degrees.
  • Add this information to the tables (one table for each triangle).
ANGLEOPPOSITE SIDE
$\, D = 27^\circ$$d = 23$
$\, E = \,?$$\,e = \,?$
$\, F = \color{red}{36.3^\circ}$$\,f = 30$


ANGLEOPPOSITE SIDE
$\, D = 27^\circ$$d = 23$
$\, E = \,?$$\,e = \,?$
$\, F = \color{red}{143.7^\circ}$$\,f = 30$
  • Compute the remaining angles: $$ \begin{gather} 180^\circ - 27^\circ - 36.3^\circ = \color{red}{116.7^\circ}\cr\cr 180^\circ - 27^\circ - 143.7^\circ = \color{red}{9.3^\circ}\cr\cr \end{gather} $$
  • Add this information to the tables.
ANGLEOPPOSITE SIDE
$\, D = 27^\circ$$d = 23$
$\, E = \color{red}{116.7^\circ}$$\,e = \,?$
$\, F = 36.3^\circ$$\,f = 30$


ANGLEOPPOSITE SIDE
$\, D = 27^\circ$$d = 23$
$\, E = \color{red}{9.3^\circ}$$\,e = \,?$
$\, F = 143.7^\circ$$\,f = 30$
  • Use the Law of Sines again to compute the remaining side: $$ \begin{gather} \frac{\sin 27^\circ}{23} = \frac{\sin 116.7^\circ}{e}\cr\cr e = \frac{23\sin 116.7^\circ}{\sin 27^\circ} \approx \color{red}{45.26}\cr\cr\cr\cr \frac{\sin 27^\circ}{23} = \frac{\sin 9.3^\circ}{e}\cr\cr e = \frac{23\sin 9.3^\circ}{\sin 27^\circ} \approx \color{red}{8.19} \end{gather} $$
  • Add this information to the tables.
ANGLEOPPOSITE SIDE
$\, D = 27^\circ$$d = 23$
$\, E = 116.7^\circ$$\,e = \color{red}{45.26}$
$\, F = 36.3^\circ$$\,f = 30$


ANGLEOPPOSITE SIDE
$\, D = 27^\circ$$d = 23$
$\, E = 9.3^\circ$$\,e = \color{red}{8.19}$
$\, F = 143.7^\circ$$\,f = 30$
  • CHECK your results by making sure the ratios in the Law of Sines are all the same: $$ \frac{\sin 27^\circ}{23} = \frac{\sin 116.7^\circ}{45.26} = \frac{\sin 36.3^\circ}{30} = 0.0197 \qquad \text{Check!} $$ $$ \frac{\sin 27^\circ}{23} = \frac{\sin 9.3^\circ}{8.19} = \frac{\sin 143.7^\circ}{30} = 0.0197 \qquad \text{Check!} $$
You can use WolframAlpha to do the computations.

For example, you can type:

23 sin(116.7 deg) / sin(27 deg)

to compute: $$ \frac{23\sin 116.7^\circ}{\sin 27^\circ} $$
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Law of Cosines
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11
12 13 14 15 16 17 18 19 20 21 22
AVAILABLE MASTERED IN PROGRESS

(MAX is 22; there are 22 different problem types.)
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