Summary: Solving Triangles (All Types)

To solve a triangle means to find all the angles and all the side lengths.
This section summarizes information related to solving triangles,
and presents ‘safe’ solution approaches for all triangle types.

Complete discussions are offered in the sections listed below; consult them as needed.

Notation

Capital letters are used to denote angles/vertices.
Corresponding lowercase letters are used to denote lengths of opposite sides.
Thus:
  • angle $\,D\,$ is opposite side $\,d\,$
  • angle $\,E\,$ is opposite side $\,e\,$
  • angle $\,F\,$ is opposite side $\,f\,$

Easy-to-use Triangle Solver

If you just need to solve a triangle reliably, here's an easy-to-use triangle solver.
It can also be used to check your work.

Basic Tools/Ideas Useful for Solving Triangles

In any triangle: Law of Sines Law of Cosines Special Tools for Right Triangles Computational Considerations

GOOD HABIT:   When solving triangles, select tools that use exact values, whenever possible.

For example, suppose you're solving the right triangle shown at right.
Green numbers are given (exact) values ($\,\color{green}{2}\,$ and $\,\color{green}{39^\circ}\,$).
Black numbers are computed exact values ($\,51^\circ\,$).
Red numbers are computed approximate values ($\,\color{red}{2.574}\,$).

To compute $\,b\,$, you should use $$ \tan 39^\circ = \frac{b}{2}\qquad \text{ or } \qquad \tan 51^\circ = \frac{2}{b} $$ since they use only exact values.

You should not use $$ \cos 51^\circ = \frac{b}{\color{red}{2.574}}\qquad \text{ or }\qquad \sin 39^\circ = \frac{b}{\color{red}{2.574}}\qquad \text{ or }\qquad b^2 + 2^2 = (\color{red}{2.574})^2 $$ since they use computed approximate values, which introduce more computational error.

GOOD HABIT:   Use the full accuracy of your calculator, whenever possible.
For example, if you've computed $\,2.574...\,$ and must use it later on,
then use your calculator features (e.g., memory, recent values) to use all the calculator digits.
Don't just type in the four digits $\,2.574\,$!


Special Triangles

Be on the lookout for these special triangles.
When you recognize them, you should be able to fill in values from memory (no computations).



$\,30^\circ\,$-$\,60^\circ\,$-$\,90^\circ\,$ triangle
shortest side:opposite $\,30^\circ\,$ angle
hypotenuse:twice the shortest side
medium side:$\,\sqrt{3}\,$ times the shortest side


$\,45^\circ\,$-$\,45^\circ\,$-$\,90^\circ\,$ triangle

hypotenuse is
$\,\sqrt{2}\,$ times the length of the leg

Solving Triangles: What Can Go Wrong?

The following example illustrates what can go wrong when solving triangles, if good habits are not followed.

Consider a triangle with sides of length $\,3\,$, $\,4\,$ and $\,6\,$.
Solve the triangle.

Faulty Solution:
  • Let $\,D,\, E,\, F\,$ be the angles across from the sides of length $\,3,\, 4,\, 6\,$.
  • Note that $\,6 < 3 + 4\,$. By SSS, the triangle is unique.
  • By the Law of Cosines: $$ \begin{gather} 3^2 = 4^2 + 6^2 - 2\cdot 4\cdot 6\cdot\cos D\cr\cr \cos D \approx 0.8958 \cr\cr D = \arccos(\cos D) \approx 26.38^\circ \end{gather} $$
  • By the Law of Sines: $$ \begin{gather} \frac{\sin 26.38^\circ}{3} = \frac{\sin F}{6}\cr\cr \sin F = \frac{6\sin 26.38^\circ}{3} \approx 0.8886\cr\cr F = \arcsin(0.8886) \approx 62.70^\circ \end{gather} $$
  • $\,E = 180^\circ - 26.38^\circ - 62.70^\circ = 90.92^\circ$
  • Summary:
    ANGLESOPPOSITE SIDES
    $D = 26.38^\circ$$d = 3$
    $E = 90.92^\circ$$e = 4$
    $F = 62.70^\circ$$f = 6$
This is an ‘impossible triangle’!
  • The longest side is not opposite the biggest angle.
  • The Law of Sines ratios are not all equal: $$ \frac{\sin 26.38^\circ}{3} \approx 0.1481 \qquad \frac{\sin 90.92^\circ}{4} \approx 0.2500 $$
An application of just one of the two recommended ‘good habits’ would have prevented the impossible triangle. Here, both good habits are applied:

Correct Solution:
  • Let $\,D,\, E,\, F\,$ be the angles across from the sides of length $\,3,\, 4,\, 6\,$.
  • Note that $\,6 < 3 + 4\,$. By SSS, the triangle is unique.
  • Use the Law of Cosines with the longest side: $$ \begin{gather} 6^2 = 3^2 + 4^2 - 2\cdot 3\cdot 4\cdot\cos F\cr\cr \cos F \approx -0.4583 \cr\cr F = \arccos(\cos F) \approx 117.3^\circ \end{gather} $$
  • Use the Law of Sines with the shortest side: $$ \begin{gather} \frac{\sin 117.3^\circ}{6} = \frac{\sin D}{3}\cr\cr \sin D = \frac{3\sin 117.3^\circ}{6} \approx 0.4443\cr\cr D = \arcsin(0.4443) \approx 26.38^\circ \end{gather} $$
  • $\,E = 180^\circ - 117.3^\circ - 26.38^\circ = 36.32^\circ$
  • Summary:
    ANGLESOPPOSITE SIDES
    $D = 26.38^\circ$$d = 3$
    $E = 36.32^\circ$$e = 4$
    $F = 117.3^\circ$$f = 6$
This is the correct triangle!

Note: At left, the corrections would be: $$\begin{gather} F = 180^\circ - \arcsin(0.8886) \approx 117.3^\circ\cr\cr E = 180^\circ - 26.38^\circ - 117.3^\circ = 36.32^\circ \end{gather} $$

Congruence Theorems

CONGRUENCE THEOREMS
A unique triangle is determined by each of these configurations:
SAS
side-angle-side
(two sides and an included angle)
SSS
side-side-side
(three sides)
AAS/SAA
angle-angle-side (or, side-angle-angle)
(two angles and a non-included side)
ASA
angle-side-angle
(two angles and an included side)

If $$\begin{gather} e > 0\cr f > 0\cr 0^\circ < D < 180^\circ \end{gather} $$ then there exists a unique triangle having the two sides
and included angle.

Name the sides so that: $$0 < f \le e \le d$$ Then, if $$d < e + f$$ the triangle exists and is unique.

That is:
the longest side (if there is one)
must be strictly less than
the sum of the two other sides.

If $$ \begin{gather} 0^\circ < D < 180^\circ\cr 0^\circ < E < 180^\circ\cr D + E < 180^\circ\cr d > 0 \end{gather} $$ then there exists a unique triangle having the two angles
and non-included side.

Note:
Two angles uniquely determine the third.
Three angles uniquely define the shape.
One side uniquely defines the size.

If $$ \begin{gather} 0^\circ < D < 180^\circ\cr 0^\circ < E < 180^\circ\cr D + E < 180^\circ\cr f > 0 \end{gather} $$ then there exists a unique triangle having the two angles
and included side.

Note:
Two angles uniquely determine the third.
Three angles uniquely define the shape.
One side uniquely defines the size.

‘SAFE’ Approaches for Solving Triangles

‘SAFE’ TRIANGLE-SOLVING STRATEGIES
By using the guidelines here, you will avoid a mistake leading to an ‘impossible’ triangle.
When it is safe to do so, the Law of Sines is preferred over the Law of Cosines, since it requires fewer computations.
Once two angles are known, find the third angle using the fact that the angles sum to $\,180^\circ\,$.
SAS

  1. Use the Law of Cosines to find $\,d\,$.
  2. Use the Law of Sines with the shorter of $\,e\,$ or $\,f\,$.
    Why?
    A triangle can have at most one obtuse angle.
    Therefore, at most one of $\,E\,$ or $\,F\,$ is obtuse.
    Whichever is opposite the shorter side is guaranteed to be acute. For acute angles: $$D = \arcsin(\sin D)$$
SSS

  1. Let $\,d\,$ denote the longest side (if there is one).
  2. Use the Law of Cosines to find $\,D\,$. If there is an obtuse angle, the Law of Cosines will find it correctly, and the remaining two angles will be acute.
  3. Use the Law of Sines to find a second angle.
AAS/SAA

  1. Use the angle sum property to find the third angle.
  2. Use the Law of Sines twice to find the remaining two sides.
Note:
There is no danger
in using the Law of Sines
when the unknown is a side.
ASA

  1. Use the angle sum property to find the third angle.
  2. Use the Law of Sines twice to find the remaining two sides.
Note:
There is no danger
in using the Law of Sines
when the unknown is a side.


OTHER TRIANGLE CONFIGURATIONS
AAA
  • Infinitely many triangles have the same three angles. Therefore, you cannot ‘solve’ an AAA triangle.
  • Triangles with the same angles are called similar triangles.
  • Similar triangles have the same shape, but not necessarily the same size.
  • Specifying the length of just one side determines the size.
    This is the justification for the AAS and ASA congruence theorems.
SSA/ASS
  • 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.
  • The cut-off points for different behaviors are summarized below.
    See Given Two Sides and a non-Included Angle, How Many Triangles?
    for details.
  • You don't need to (and shouldn't) memorize the table below!
    The mathematics will tell you which situation you're in (zero, one, or two triangles), providing you do the mathematics correctly!
Let: $$\begin{gather} a > 0\cr\cr b > 0\cr\cr 0^\circ < \theta < 180^\circ \end{gather} $$ Note:
the side of length $\,a\,$
is opposite the known angle $\,\theta\,$
$\theta < 90^\circ$
($\,\theta\,$ is acute)
$0 < a < b\sin\theta$ no triangle
$a = b\sin\theta$ exactly one right triangle
$b\sin\theta < a < b$ two different triangles
$a = b$ exactly one isosceles triangle
$a > b$ exactly one triangle
$\theta \ge 90^\circ$
($\,\theta\,$ is a right angle or obtuse)
$0 < a \le b$ no triangle
$a > b$ exactly one triangle


SOLVING SSA CONFIGURATIONS: TWO APPROACHES
Each step is illustrated with this example:
Suppose a triangle has a $\,50^\circ\,$ angle opposite a side of length $\,6\,$.
A second side has length $\,7\,$. Solve the triangle.

Start by defining notation:
Let $\,E\,$ be the angle opposite the side of length $\,7\,$.
Let $\,F,\,f\,$ be the remaining unknown angle and opposite side.
USE THE LAW OF SINES:
JUST BE CAREFUL
USE THE LAW OF COSINES:
SOLVE A QUADRATIC EQUATION
  1. Recognize the SSA configuration.
    Be aware of the potential for zero, one, or two triangles.
  2. Use the Law of Sines to get both potential angles: $$ \begin{gather} \frac{\sin 50^\circ}{6} = \frac{\sin E}{7}\cr\cr \sin E = \frac{7\sin 50^\circ}{6} \approx 0.8937 \cr\cr E_1 = \arcsin(0.8937) \approx 63.34^\circ\cr E_2 = 180^\circ - 63.34^\circ = 116.66^\circ \end{gather} $$
  3. Get both potential third angles: $$ \begin{gather} F_1 = 180^\circ - 50^\circ - 63.34^\circ = 66.66^\circ\cr F_2 = 180^\circ - 50^\circ - 116.66^\circ = 13.34^\circ \end{gather} $$ (Of course, discard a situation that results in a nonpositive angle.)
  4. Use the Law of Sines to find the remaining side(s): $$ \begin{gather} \frac{\sin 50^\circ}{6} = \frac{\sin 66.66^\circ}{f_1}\qquad \text{so}\qquad f_1 \approx 7.19\cr\cr \frac{\sin 50^\circ}{6} = \frac{\sin 13.34^\circ}{f_2}\qquad \text{so}\qquad f_2 \approx 1.81 \end{gather} $$
Compare the results.

Different approaches may yield slightly different results.

Greater accuracy at intermediate steps
will help to minimize differences.
  1. Recognize the SSA configuration.
    Be aware of the potential for zero, one, or two triangles.
  2. Use the Law of Cosines starting with the side across from the known angle: $$ 6^2 = 7^2 + f^2 - 2\cdot 7\cdot f\cos 50^\circ $$
  3. Solve the quadratic equation.
    It may give zero, one, or two solutions for $\,f\,$: $$ \begin{gather} f^2 - (14\cos 50^\circ)f + 13 = 0\cr\cr f = \frac{14\cos 50^\circ \pm \sqrt{14^2\cos^2 50^\circ - 4(13)}}{2}\cr\cr f_1 \approx 7.19\cr f_2 \approx 1.81 \end{gather} $$
  4. Use the Law of Sines on each configuration.
    Choose the shortest available side for each configuration.
    Note:
    first configuration: $\,e_1 = 7\,$ and $\,f_1 = 7.19\,$, so $\,\color{red}{7} < 7.191\,$
    second configuration: $\,e_2 = 7\,$ and $\,f_2 = 1.81\,$, so $\,\color{green}{1.81} < 7\,$ $$ \begin{gather} \frac{\sin 50^\circ}{6} = \frac{\sin E_1}{\color{red}{7}}\qquad \text{so}\qquad E_1 \approx 63.34^\circ\cr\cr \frac{\sin 50^\circ}{6} = \frac{\sin F_2}{\color{green}{1.81}}\qquad \text{so}\qquad F_2 \approx 13.36^\circ \end{gather} $$
  5. Compute the remaining angles: $$ \begin{gather} F_1 = 180^\circ - 50^\circ - 63.34^\circ = 66.66^\circ\cr E_2 = 180^\circ - 50^\circ - 13.36^\circ = 116.64^\circ \end{gather} $$
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
Verifying Trigonometric Identities
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.

Although only a few decimal places are displayed in problem solutions, additional accuracy is used in intermediate calculations.
PROBLEM TYPES:
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14 15 16 17 18 19 20 21 22 23 24 25  
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(MAX is 25; there are 25 different problem types.)
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