5.12: General Solutions of Triangles
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 \begin{align*}\frac{1}{2}bh\end{align*}
"Do you know how to find the lengths of sides of the triangle and the area?" she asks.
Read on, and at the end of this Concept, you'll be able to answer your sister's question.
Watch This
James Sousa: The Law of Sines: The Basics
Guidance
In the previous sections we have discussed a number of methods for finding a missing side or angle in a triangle. Previously, we only knew how to do this in right triangles, but now we know how to find missing sides and angles in oblique triangles as well. By combining all of the methods we’ve learned up until this point, it is possible for us to find all missing sides and angles in any triangle we are given.
Below is a chart summarizing the triangle techniques that we have learned up to this point. 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 nonincluded side (AAS)  Law of Sines  The other nonincluded side 
Oblique  2 angles and the included side (ASA)  Law of Sines  Either of the nonincluded 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 
Example A
In \begin{align*}\triangle ABC, a = 12, b = 13, c = 8\end{align*}
Solution: 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 \begin{align*}\angle{A}, \angle{B}\end{align*}
\begin{align*}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{align*}
Now, we will find \begin{align*}\angle{B}\end{align*}
\begin{align*}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{align*}
We can now quickly find \begin{align*}\angle{C}\end{align*}
Example B
In triangle \begin{align*}DEF, d = 43, e = 37\end{align*}
Solution: 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 \begin{align*}f,\angle{D}\end{align*}
\begin{align*}f^2 & = 43^2 + 37^2  2(43)(37) \cos 124 \\ f^2 & = 4997.351819 \\ f& \approx 70.7\end{align*}
Now that we know \begin{align*}f\end{align*}
\begin{align*}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{align*}
To find \begin{align*}\angle E\end{align*}
Example C
In triangle \begin{align*}ABC, A = 43^\circ, B = 82^\circ\end{align*}
Solution: 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 \begin{align*}\angle C\end{align*}
Now that we know \begin{align*}\angle{C}\end{align*}
\begin{align*}\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} && \qquad \ b = \frac{10.3 \sin 82}{\sin 55} \\ a & = 8.6 && \qquad \ b = 12.5\end{align*}
Guided Practice
1. 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.
\begin{align*}A = 69^\circ, B = 12^\circ, a = 22.3\end{align*}
2. 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.
\begin{align*}a = 1.4, b = 2.3, C = 58^\circ\end{align*}
3. 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.
\begin{align*}a = 3.3, b = 6.1, c = 4.8\end{align*}
Solutions:
1. AAS, Law of Sines, one solution
2. SAS, Law of Cosines, one solution
3. SSS, Law of Cosines, one solution
Concept Problem Solution
Since you know that two of the angles are \begin{align*}23^\circ\end{align*}
\begin{align*} \frac{\sin A}{a} = \frac{\sin B}{b}\\ \frac{\sin 23^\circ}{a} = \frac{\sin 129^\circ}{4}\\ a = \frac{4\sin 23^\circ}{\sin 129^\circ} = \frac{1.56}{.777}\\ a \approx 2\\ \end{align*}
And repeating the process for the third side:
\begin{align*} \frac{\sin A}{a} = \frac{\sin C}{c}\\ \frac{\sin 23^\circ}{2} = \frac{\sin 28^\circ}{c}\\ c = \frac{2\sin 28^\circ}{\sin 23^\circ} = \frac{.939}{.781}\\ c \approx 1.2\\ \end{align*}
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{align*} K = \frac{1}{2}bc\sin A\\ K = \frac{1}{2}(4)(1.2)\sin 23^\circ\\ K = \frac{1}{2}(4)(1.2)(.391)\\ K \approx .9384 \end{align*}
Explore More
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.

\begin{align*}a=3, b=4, C=71^\circ\end{align*}
a=3,b=4,C=71∘ , find \begin{align*}c\end{align*}c . 
\begin{align*}a=8, b=7, c=9\end{align*}
a=8,b=7,c=9 , find \begin{align*}A\end{align*}.  \begin{align*}A=135^\circ, B=12^\circ, c=100\end{align*}, find \begin{align*}a\end{align*}.
 \begin{align*}a=12, b=10, A=80^\circ\end{align*}, find \begin{align*}c\end{align*}.
 \begin{align*}A=50^\circ, B=87^\circ, a=13\end{align*}, find \begin{align*}b\end{align*}.
 In \begin{align*}\triangle ABC\end{align*}, \begin{align*}a=15, b=19, c=20\end{align*}. Solve the triangle.
 In \begin{align*}\triangle DEF\end{align*}, \begin{align*}d=12, E=39^\circ, f=17\end{align*}. Solve the triangle.
 In \begin{align*}\triangle PQR\end{align*}, \begin{align*}P=115^\circ, Q=30^\circ, q=10\end{align*}. Solve the triangle.
 In \begin{align*}\triangle MNL\end{align*}, \begin{align*}m=5, n=9, L=20^\circ\end{align*}. Solve the triangle.
 In \begin{align*}\triangle SEV\end{align*}, \begin{align*}S=50^\circ, E=44^\circ, s=12\end{align*}. Solve the triangle.
 In \begin{align*}\triangle KTS\end{align*}, \begin{align*}k=6, t=15, S=68^\circ\end{align*}. Solve the triangle.
 In \begin{align*}\triangle WRS\end{align*}, \begin{align*}w=3, r=5, s=6\end{align*}. Solve the triangle.
 In \begin{align*}\triangle DLP\end{align*}, \begin{align*}D=52^\circ, L=110^\circ, p=8\end{align*}. Solve the triangle.
 In \begin{align*}\triangle XYZ\end{align*}, \begin{align*}x=10, y=12, z=9\end{align*}. Solve the triangle.
 In \begin{align*}\triangle AMF\end{align*}, \begin{align*}A=99^\circ, m=15, f=16\end{align*}. Solve the triangle.
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 , where is the angle across from side .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.Image Attributions
Description
Learning Objectives
In this Concept we'll use the Pythagorean Theorem, trigonometry functions, the Law of Sines, and the Law of Cosines to solve various triangles. Our focus will be on understanding when it is appropriate to use each method as well as how to apply the methods above in realworld and applied problems.
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Date Created:
Sep 26, 2012Last Modified:
Feb 26, 2015Vocabulary
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