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# 8.7: Extension: Laws of Sines and Cosines

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Identify and use the Law of Sines and Cosines.

In this chapter, we have only applied the trigonometric ratios to right triangles. However, you can extend what we know about these ratios and derive the Law of Sines and the Law of Cosines. Both of these laws can be used with any type of triangle to find any angle or side within it. That means we can find the sine, cosine and tangent of angle that are greater than 90\begin{align*}90^\circ\end{align*}, such as the obtuse angle in an obtuse triangle.

## Law of Sines

Law of Sines: If ABC\begin{align*}\triangle ABC\end{align*} has sides of length, a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*}, then sinAa=sinBb=sinCc\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\end{align*}.

Looking at a triangle, the lengths a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*} are opposite the angles of the same letter. Let’s use the Law of Sines on a couple of examples.

We will save the proof for a later course.

Example 1: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth.

Solution: First, to find mA\begin{align*}m \angle A\end{align*}, we can use the Triangle Sum Theorem.

mA+85+38mA=180=57\begin{align*}m \angle A + 85^\circ + 38^\circ & = 180^\circ\\ m \angle A & = 57^\circ\end{align*}

Now, use the Law of Sines to set up ratios for a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}.

sin57a=sin85b=sin3812\begin{align*}\frac{\sin 57^\circ}{a} = \frac{\sin 85^\circ}{b} = \frac{\sin 38^\circ}{12}\end{align*}

sin57aasin38a=sin3812=12sin57=12sin57sin3816.4sin85b=sin3812bsin38=12sin85  b=12sin85sin3819.4\begin{align*}\frac{\sin 57^\circ}{a} & = \frac{\sin 38^\circ}{12} && \quad \frac{\sin 85^\circ}{b} = \frac{\sin 38^\circ}{12}\\ a \cdot \sin 38^\circ & = 12 \cdot \sin 57^\circ && b \cdot \sin 38^\circ = 12 \cdot \sin 85^\circ\\ a & = \frac{12 \cdot \sin 57^\circ}{\sin 38^\circ} \approx 16.4 && \qquad \quad \ \ b = \frac{12 \cdot \sin 85^\circ}{\sin 38^\circ} \approx 19.4\end{align*}

Example 2: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth.

Solution: Set up the ratio for B\begin{align*}\angle B\end{align*} using Law of Sines.

sin952727sinBsinB=sinB16=16sin95=16sin9527sin1(16sin9527)=36.2\begin{align*}\frac{\sin 95^\circ}{27} & = \frac{\sin B}{16}\\ 27 \cdot \sin B & = 16 \cdot \sin 95^\circ\\ \sin B & = \frac{16 \cdot \sin 95^\circ}{27} \rightarrow \sin^{-1} \left ( \frac{16 \cdot \sin 95^\circ}{27} \right ) = 36.2^\circ\end{align*}

To find mC\begin{align*}m \angle C\end{align*} use the Triangle Sum Theorem. mC+95+36.2=180mC=48.8\begin{align*}m \angle C + 95 ^\circ + 36.2^\circ = 180^\circ \rightarrow m \angle C = 48.8^\circ\end{align*}

To find c\begin{align*}c\end{align*}, use the Law of Sines again. sin9527=sin48.8c\begin{align*}\frac{\sin 95^\circ}{27} = \frac{\sin 48.8^\circ}{c}\end{align*}

csin95c=27sin48.8=27sin48.8sin9520.4\begin{align*}c \cdot \sin 95^\circ & = 27 \cdot \sin 48.8^\circ\\ c & = \frac{27 \cdot \sin 48.8^\circ}{\sin 95^\circ} \approx 20.4\end{align*}

## Law of Cosines

Law of Cosines: If ABC\begin{align*}\triangle ABC\end{align*} has sides of length a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*}, then a2=b2+c22bccosA\begin{align*}a^2 = b^2 + c^2 - 2bc \cos A\end{align*}

b2c2=a2+c22ac cosB=a2+b22ab cosC\begin{align*}b^2 & = a^2 + c^2 - 2ac \ \cos B\\ c^2 & = a^2 + b^2 - 2ab \ \cos C\end{align*}

Even though there are three formulas, they are all very similar. First, notice that whatever angle is in the cosine, the opposite side is on the other side of the equal sign.

Example 3: Solve the triangle using Law of Cosines. Round your answers to the nearest hundredth.

Solution: Use the second equation to solve for B\begin{align*}\angle B\end{align*}.

b2b2b2b=262+1822(26)(18)cos26=1000936cos26=158.728812.60\begin{align*}b^2 & = 26^2 + 18^2 - 2(26)(18) \cos 26^\circ\\ b^2 & = 1000 - 936 \cos 26^\circ\\ b^2 & = 158.7288\\ b & \approx 12.60\end{align*}

To find mA\begin{align*}m \angle A\end{align*} or mC\begin{align*}m \angle C\end{align*}, you can use either the Law of Sines or Law of Cosines. Let’s use the Law of Sines.

sin2612.60=sinA1812.60sinAsinA=18sin26=18sin2612.60\begin{align*}\frac{\sin 26^\circ}{12.60} = \frac{\sin A}{18}&&\\ && 12.60 \cdot \sin A & = 18 \cdot \sin 26^\circ\\ && \sin A & = \frac{18 \cdot \sin 26^\circ}{12.60}\end{align*}

sin1(18sin2612.60)38.77\begin{align*}\sin^{-1} \left ( \frac{18 \cdot \sin 26^\circ}{12.60} \right ) \approx 38.77^\circ\end{align*} To find mC\begin{align*}m \angle C\end{align*}, use the Triangle Sum Theorem.

26+38.77+mCmC=180=115.23\begin{align*}26^\circ + 38.77^\circ + m \angle C & = 180^\circ\\ m \angle C & =115.23^\circ\end{align*}

Unlike the previous sections in this chapter, with the Laws of Sines and Cosines, we have been using values that we have found to find other values. With these two laws, you have to use values that are not given. Just keep in mind to always wait until the very last step to put anything into your calculator. This will ensure that you have the most accurate answer.

Example 4: Solve the triangle. Round your answers to the nearest hundredth.

Solution: When you are given only the sides, you have to use the Law of Cosines to find one angle and then you can use the Law of Sines to find another.

152225104310431232=222+2822(22)(28)cosA=12681232cosA=1232cosA=cosAcos1(10431232)32.16\begin{align*}15^2 & = 22^2 + 28^2 - 2(22)(28) \cos A\\ 225 & = 1268 - 1232 \cos A\\ -1043 & = -1232 \cos A\\ \frac{-1043}{-1232} & = \cos A \rightarrow \cos^{-1} \left (\frac{1043}{1232} \right ) \approx 32.16^\circ\end{align*}

Now that we have an angle and its opposite side, we can use the Law of Sines.

sin32.1615=sinB2215sinBsinB=22sin32.16=22sin32.1615\begin{align*}\frac{\sin 32.16^\circ}{15} = \frac{\sin B}{22}&&\\ &&15 \cdot \sin B & = 22 \cdot \sin 32.16^\circ\\ &&\sin B & = \frac{22 \cdot \sin 32.16^\circ}{15}\end{align*}

sin1(22sin32.1615)51.32\begin{align*}\sin^{-1} \left ( \frac{22 \cdot \sin 32.16^\circ}{15} \right ) \approx 51.32^\circ\end{align*} To find mC\begin{align*}m \angle C\end{align*}, use the Triangle Sum Theorem.

32.16+51.32+mCmC=180=96.52\begin{align*}32.16^\circ + 51.32^\circ + m \angle C &= 180^\circ\\ m \angle C &= 96.52^\circ\end{align*}

## To Summarize

Use Law of Sines when given:

• An angle and its opposite side.
• Any two angles and one side.
• Two sides and the non-included angle.

Use Law of Cosines when given:

• Two sides and the included angle.
• All three sides.

## Review Questions

Use the Law of Sines or Cosines to solve ABC\begin{align*}\triangle ABC\end{align*}. If you are not given a picture, draw one. Round all decimal answers to the nearest tenth.

1. mA=74,mB=11,BC=16\begin{align*}m \angle A = 74^\circ, m \angle B = 11^\circ, BC = 16\end{align*}
2. mA=64,AB=29,AC=34\begin{align*}m∠A = 64^\circ, AB = 29, AC = 34\end{align*}
3. mC=133,mB=25,AB=48\begin{align*}m \angle C = 133^\circ, m \angle B = 25^\circ, AB=48\end{align*}

Use the Law of Sines to solve ABC\begin{align*}\triangle ABC\end{align*} below.

1. mA=20,AB=12,BC=5\begin{align*} m\angle A = 20^\circ, AB = 12, BC = 5\end{align*}

Recall that when we learned how to prove that triangles were congruent we determined that SSA (two sides and an angle not included) did not determine a unique triangle. When we are using the Law of Sines to solve a triangle and we are given two sides and the angle not included, we may have two possible triangles. Problem 14 illustrates this.

1. Let’s say we have ABC\begin{align*} \triangle ABC \end{align*} as we did in problem 13. In problem 13 you were given two sides and the not included angle. This time, you have two angles and the side between them (ASA). Solve the triangle given that mA=20,mC=125,AC=8.4\begin{align*}m \angle A = 20^\circ, m \angle C = 125^\circ, AC = 8.4\end{align*}
2. Does the triangle that you found in problem 14 meet the requirements of the given information in problem 13? How are the two different mC\begin{align*}m \angle C\end{align*} related? Draw the two possible triangles overlapping to visualize this relationship.

It is beyond the scope of this text to determine when there will be two possible triangles, but the concept of the possibility is something worth noting at this time.

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