# 4.2.5: General Solutions of Triangles

- Page ID
- 4167

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Application of pythagorean theorem, trigonometric functions, and laws of sines/cosines.

While talking with your little sister one day, the conversation turns to shapes. Your sister is only in junior high school, so while she knows some things about right triangles, such as the Pythagorean Theorem, she doesn't know anything about other types of triangles. You show her an example of an oblique triangle by drawing this on a piece of paper:

Fascinated, she tells you that she knows how to calculate the area of a triangle using the familiar formula \(\dfrac{1}{2} bh\) and the lengths of sides if the triangle is a right triangle, but that she can't use the formulas on the triangle you just drew.

"Do you know how to find the lengths of sides of the triangle and the area?" she asks.

## Finding Solutions for Triangles

Finding the sides, angles, and area for right triangles is often learned in Algebra and/or Geometry. However, it is common to learn how to determine this information in non-right triangles in Trigonometry.

Below is a chart summarizing

common

triangle techniques. This chart describes the type of triangle (either right or oblique), the given information, the appropriate technique to use, and what we can find using each technique.

Type of Triangle: |
Given Information: |
Technique: |
What we can find: |
---|---|---|---|

Right | Two sides | Pythagorean Theorem | Third side |

Right | One angle and one side | Trigonometric ratios | Either of the other two sides |

Right | Two sides | Trigonometric ratios | Either of the other two angles |

Oblique | 2 angles and a non-included side (AAS) | Law of Sines |
The other non-included side |

Oblique | 2 angles and the included side (ASA) | Law of Sines | Either of the non-included sides |

Oblique | 2 sides and the angle opposite one of those sides (SSA) – Ambiguous case | Law of Sines | The angle opposite the other side (can yield no, one, or two solutions) |

Oblique | 2 sides and the included angle (SAS) | Law of Cosines |
The third side |

Oblique | 3 sides | Law of Cosines | Any of the three |

angles |

#### Solving Triangles

1. In \(\Delta ABC\), \(a=12\), \(b=13\), \(c=8\). Solve the triangle.

Since we are given all three sides in the triangle, we can use the Law of Cosines. Before we can solve the triangle, it is important to know what information we are missing. In this case, we do not know any of the angles, so we are solving for \(\angle A\), \(\angle B\), and \(\angle C\). We will begin by finding \(\angle A\):

\(\begin{aligned} 12^2 &=8^2+13^2−2(8)(13)\cos A\\ 144 &=233−208\cos A\\ −89 &=−208\cos A\\ 0.4278846154&=\cos A\\ 64.7&\approx \angle A \end{aligned}\)

Now, we will find \angle B by using the Law of Cosines. Keep in mind that you can now also use the Law of Sines to find \(\angle B\). Use whatever method you feel more comfortable with.

\(\begin{aligned} 13^2 &=8^2+12^2−2(8)(12) \cos B\\ 169 &=208−192\cos B\\ −39 &=−192\cos B\\ 0.2031 &=\cos B\\ 78.3^{\circ} & \approx \angle B \end{aligned}\)

We can now quickly find \(\angle C\) by using the Triangle Sum Theorem,

\(180^{\circ} −64.7^{\circ} −78.3^{\circ} =37^{\circ}\)

2. In triangle \(DEF\), \(d=43\), \(e=37\), and \(\angle F=124^{\circ} \). Solve the triangle.

In this triangle, we have the SAS case because we know two sides and the included angle. This means that we can use the Law of Cosines to solve the triangle. In order to solve this triangle, we need to find side \(f\), \(\angle D\), and \(\angle E\). First, we will need to find side \(f\) using the Law of Cosines.

\(\begin{aligned}

f^{2} &=43^{2}+37^{2}-2(43)(37) \cos 124 \\

f^{2} &=4997.351819 \\

f & \approx 70.7

\end{aligned}\)

Now that we know \(f\), we know all three sides of the triangle. This means that we can use the Law of Cosines to find either \(\angle D\) or \(\angle E\). We will find \(\angle D\) first.

\(\begin{aligned}

43^{2} &=70.7^{2}+37^{2}-2(70.7)(37) \cos D \\

1849 &=6367.49-5231.8 \cos D \\

-4518.49 &=-5231.8 \cos D \\

0.863658779 &=\cos D \\

30.3^{\circ} & \approx \angle D

\end{aligned}\)

To find \(\angle E\), we need only to use the Triangle Sum Theorem, \(\angle E=180−(124+30.3)=25.7^{\circ} \).

3. In triangle \(ABC\), \(A=43^{\circ} \), \(B=82^{\circ} \), and \(c=10.3\). Solve the triangle.

This is an example of the ASA case, which means that we can use the Law of Sines to solve the triangle. In order to use the Law of Sines, we must first know \(\angle C\), which we can find using the Triangle Sum Theorem, \(\angle C=180^{\circ} −(43^{\circ} +82^{\circ} )=55^{\circ} \).

Now that we know \(\angle C\), we can use the Law of Sines to find either side a or side \(b\).

\(\begin{aligned}

\frac{\sin 55}{10.3} & =\frac{\sin 43}{a} & \frac{\sin 55}{10.3} & =\frac{\sin 82}{b} \\

a & =\frac{10.3 \sin 43}{\sin 55} & b & =\frac{10.3 \sin 82}{\sin 55} \\

a & =8.6 & b & =12.5

\end{aligned}\)

Earlier, you were asked how you might help your sister find the lengths of the sides and the area of a non-right triangle.

**Solution**

Since you know that two of the angles are \(23^{\circ} \) and \(28^{\circ} \), the third angle in the triangle must be \(180^{\circ} −23^{\circ} −28^{\circ} =129^{\circ} \). Using these angles and the knowledge that one of the sides has a length of 4, you can solve for the lengths of the other two sides using the Law of Sines:

\(\begin{aligned} \dfrac{\sin A}{a}&=\dfrac{\sin B}{b} \\ \dfrac{\sin 23^{\circ} }{a}&=\dfrac{\sin 129^{\circ} }{4} \\ a&=\dfrac{4 \sin 23^{\circ} }{\sin 129^{\circ} }=\dfrac{1.56}{.777} \\ a&\approx 2 \end{aligned}\)

And repeating the process for the third side:

\(\begin{aligned} \dfrac{\sin A}{a}&=\dfrac{\sin C}{c} \\ \dfrac{\sin 23^{\circ} }{2}&= \dfrac{\sin 28^{\circ} }{c} \\ c&=\dfrac{2\sin 28^{\circ} }{\sin 23^{\circ} }=\dfrac{.939}{.781} \\ c&\approx 1.2 \end{aligned}\)

Now you know all three angles and all three sides. You can use Heron's formula or the alternative formula for the area of a triangle to find the area:

\(\begin{aligned} K&=\dfrac{1}{2} bc\sin A \\ K&=\dfrac{1}{2} (4)(1.2)\sin 23^{\circ} \\K&=\dfrac{1}{2}(4)(1.2)(.391) \\ K&\approx .9384 \end{aligned}\)

Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have. If a triangle has no solution or two solutions, explain why.

**Solution**

\(A=69^{\circ} \), \(B=12^{\circ} \), \(a=22.3\), find \(b\)

AAS, Law of Sines, one solution

Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have. If a triangle has no solution or two solutions, explain why.

**Solution**

\(a=1.4\), \(b=2.3\), \(C=58^{\circ} \), find \(c\).

SAS, Law of Cosines, one solution

Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have. If a triangle has no solution or two solutions, explain why.

**Solution**

\(a=3.3\), \(b=6.1\), \(c=4.8\), find \(A\).

SSS, Law of Cosines, one solution

## Review

Using the information provided, decide which case you are given (SSS, SAS, AAS, ASA, or SSA), and whether you would use the Law of Sines or the Law of Cosines to find the requested side or angle. Make an approximate drawing of the triangle and label the given information. Also, state how many solutions (if any) the triangle would have.

- \(a=3\), \(b=4\), \(C=71^{\circ} \), find \(c\).
- \(a=8\), \(b=7\), \(c=9\), find \(A\).
- \(A=135^{\circ}\), \(B=12^{\circ} \), \(c=100\), find \(a\).
- \(a=12\), \(b=10\), \(A=80^{\circ} \), find \(c\).
- \(A=50^{\circ} \), \(B=87^{\circ}\), \(a=13\), find \(b\).
- In \(\Delta ABC\), \(a=15\), \(b=19\), \(c=20\). Solve the triangle.
- In \(\Delta DEF\), \(d=12\), \(E=39^{\circ}\), \(f=17\). Solve the triangle.
- In \(\Delta PQR\), \(P=115^{\circ} \), \(Q=30^{\circ} \), \(q=10\). Solve the triangle.
- In \(\Delta MNL\), \(m=5\), \(n=9\), \(L=20^{\circ} \). Solve the triangle.
- In \(\Delta SEV\), \(S=50^{\circ} \), \(E=44^{\circ} \), \(s=12\). Solve the triangle.
- In \(\Delta KTS\), \(k=6\), \(t=15\), \(S=68^{\circ} \). Solve the triangle.
- In \(\Delta WRS\), \(w=3\), \(r=5\), \(s=6\). Solve the triangle.
- In \(\Delta DLP\), \(D=52^{\circ}\), \(L=110^{\circ} \), \(p=8\). Solve the triangle.
- In \(\Delta XYZ\), \(x=10\), \(y=12\), \(z=9\). Solve the triangle.
- In \(\Delta AMF\), \(A=99^{\circ}\), \(m=1\), \(f=16\). Solve the triangle.

### Review (Answers)

To see the Review answers, open this PDF file and look for section 5.12.

## Vocabulary

Term | Definition |
---|---|

law of cosines |
The law of cosines is a rule relating the sides of a triangle to the cosine of one of its angles. The law of cosines states that \(c^2=a^2+b^2−2ab\cos C\), where \(C\) is the angle across from side \(c\). |

law of sines |
The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of another angle in the triangle to the side opposite that angle. |