law of sines homework answers

The Law of Sines

The Law of Sines (or Sine Rule ) is very useful for solving triangles:

a sin A = b sin B = c sin C

It works for any triangle:

And it says that:

When we divide side a by the sine of angle A it is equal to side b divided by the sine of angle B , and also equal to side c divided by the sine of angle C

Well, let's do the calculations for a triangle I prepared earlier:

a sin A = 8 sin(62.2°) = 8 0.885... = 9.04...

b sin B = 5 sin(33.5°) = 5 0.552... = 9.06...

c sin C = 9 sin(84.3°) = 9 0.995... = 9.04...

The answers are almost the same! (They would be exactly the same if we used perfect accuracy).

So now you can see that:

Is This Magic?

Not really, look at this general triangle and imagine it is two right-angled triangles sharing the side h :

The sine of an angle is the opposite divided by the hypotenuse, so:

a sin(B) and b sin(A) both equal h , so we get:

a sin(B) = b sin(A)

Which can be rearranged to:

a sin A = b sin B

We can follow similar steps to include c/sin(C)

How Do We Use It?

Let us see an example:

Example: Calculate side "c"

Now we use our algebra skills to rearrange and solve:

Finding an Unknown Angle

In the previous example we found an unknown side ...

... but we can also use the Law of Sines to find an unknown angle .

In this case it is best to turn the fractions upside down ( sin A/a instead of a/sin A , etc):

sin A a = sin B b = sin C c

Example: Calculate angle B

Sometimes there are two answers .

There is one very tricky thing we have to look out for:

Two possible answers.

This only happens in the " Two Sides and an Angle not between " case, and even then not always, but we have to watch out for it.

Just think "could I swing that side the other way to also make a correct answer?"

Example: Calculate angle R

The first thing to notice is that this triangle has different labels: PQR instead of ABC. But that's OK. We just use P,Q and R instead of A, B and C in The Law of Sines.

But wait! There's another angle that also has a sine equal to 0.9215...

The calculator won't tell you this but sin(112.9°) is also equal to 0.9215...

So, how do we discover the value 112.9°?

Easy ... take 67.1° away from 180°, like this:

180° − 67.1° = 112.9°

So there are two possible answers for R: 67.1° and 112.9° :

Both are possible! Each one has the 39° angle, and sides of 41 and 28.

So, always check to see whether the alternative answer makes sense.

• ... sometimes it will (like above) and there are two solutions
• ... sometimes it won't (like below) and there is one solution

For example this triangle from before.

As you can see, we can try swinging the "5.5" line around, but no other solution makes sense.

So this has only one solution.

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4.1.1: Laws of Sines and Cosines

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Solving for parts of non-right triangles using trigonometry.

Law of Sines: If \(\Delta ABC\) has sides of length, \(a\), \(b\), and \(c\), then \(\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\).

Looking at a triangle, the lengths a,b, and c are opposite the angles of the same letter.

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.

Law of Cosines: If \(\Delta ABC\) has sides of length \(a\),\(b\), and \(c\), then:

\(\begin{aligned} a^2&=b^2+c^2−2bc\cos A \\ b^2&=a^2+c^2−2ac \cos B \\ c^2&=a^2+b^2−2ab \cos C \end{aligned}\)

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.

Use Law of Cosines when given:

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

Using the Law of Sines

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

First, to find \(m\angle A\), we can use the Triangle Sum Theorem.

\(\begin{aligned} m\angle A+85^{\circ} +38^{\circ}&=180^{\circ} \\ m\angle A&=57^{\circ} \end{aligned}\)

Now, use the Law of Sines to set up ratios for \(a\) and \(b\).

\(\dfrac{\sin 57^{\circ} }{a}=\dfrac{\sin 85^{\circ} }{b}=\dfrac{\sin 38^{\circ} }{12}\)

\(\begin{aligned} \dfrac{\sin 57^{\circ}}{a} &=\dfrac{\sin 38^{\circ}}{12} & \dfrac{\sin 85^{\circ}}{b} &=\dfrac{\sin 38^{\circ}}{12} \\ a \cdot \sin 38^{\circ} &=12 \cdot \sin 57^{\circ} & b \cdot \sin 38^{\circ} &=12 \cdot \sin 85^{\circ} \\ a &=\dfrac{12 \cdot \sin 57^{\circ}}{\sin 38^{\circ}} \approx 16.4 & \quad b &=\dfrac{12 \cdot \sin 85^{\circ}}{\sin 38^{\circ}} \approx 19.4 \end{aligned}\)

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

Set up the ratio for \(\angle B\) using Law of Sines.

\(\begin{aligned} \dfrac{\sin 95^{\circ} }{27}&=\dfrac{\sin B}{16} \\ 27\cdot \sin B&=16\cdot \sin 95^{\circ} \\ \sin B&=\dfrac{16\cdot \sin 95^{\circ} }{27}\rightarrow\sin ^{−1}\left(\dfrac{16\cdot \sin 95^{\circ} }{27} \right) =36.2^{\circ} \end{aligned}\)

To find \(m\angle C\) use the Triangle Sum Theorem.

\(m\angle C+95^{\circ} +36.2^{\circ} =180^{\circ} \rightarrow m\angle C=48.8^{\circ}\)

To find \(c\), use the Law of Sines again. \(\dfrac{\sin 95^{\circ} }{27}=\dfrac{\sin 48.8^{\circ} }{c}\)

\(\begin{aligned} c\cdot \sin 95^{\circ}&=27\cdot \sin 48.8^{\circ} \\ c&=27\cdot \sin 48.8^{\circ} \sin 95^{\circ} \approx 20.4 \end{aligned}\)

Using the Law of Cosines

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

Use the second equation to solve for \(\angle B\).

\(\begin{aligned} b^2&=26^2+18^2−2(26)(18)\cos 26^{\circ} \\ b^2&=1000−936\cos 26^{\circ} \\ b^2&=158.7288 \\ b&\approx 12.60 \end{aligned}\)

To find \(m\angle A\) or \(m\angle C\), you can use either the Law of Sines or Law of Cosines . Let’s use the Law of Sines.

\(\begin{aligned} \dfrac{\sin 26^{\circ} }{12.60}&=\dfrac{\sin A}{18} \\ 12.60\cdot \sin A&=18\cdot \sin 26^{\circ} \\ \sin A&=18\cdot \sin 26^{\circ} 12.60 \end{aligned}\)

\(\sin ^{−1} \left(18\cdot \sin 26^{\circ} 12.60 \right)\approx 38.77^{\circ} \) To find \(m\angle C\), use the Triangle Sum Theorem.

\(\begin{aligned} 26^{\circ} +38.77^{\circ} +m\angle C&=180^{\circ} \\ m\angle C&=115.23^{\circ} \end{aligned}\)

Find the following angles in the triangle below. Round your answers to the nearest hundredth.

Example \(\PageIndex{1}\)

\(m\angle A\)

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.

\(\begin{aligned} 15^2&=22^2+28^2−2(22)(28)\cos A \\ 225&=1268−1232\cos A \\ −1043&=−1232\cos A \\ \dfrac{−1043}{−1232}&=\cos A \rightarrow \cos ^{−1}\left(\dfrac{1043}{1232}\right) \approx 32.16^{\circ}\end{aligned}\)

Example \(\PageIndex{2}\)

\(m\angle B\)

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

\(\begin{aligned} \dfrac{\sin 32.16^{\circ} }{15} &=\dfrac{\sin B}{22} \\ 15\cdot \sin B &=22\cdot \sin 32.16^{\circ} \\ \sin B &=\dfrac{22\cdot \sin 32.16^{\circ} }{15} \end{aligned}\)

\(\sin ^{−1}\left(\dfrac{22\cdot \sin 32.16^{\circ} }{15}\right)\approx 51.32^{\circ} \).

Example \(\PageIndex{3}\)

\(m\angle C\)

To find \(m\angle C\), use the Triangle Sum Theorem.

\(\begin{aligned} 32.16^{\circ} +51.32^{\circ} +m\angle C&=180^{\circ} \\ m\angle C&=96.52^{\circ} \end{aligned}\)

Use the Law of Sines or Cosines to solve \(\Delta ABC\). If you are not given a picture, draw one. Round all decimal answers to the nearest tenth.

• \(m\angle A=74^{\circ} \), \(m\angle B=11^{\circ} \), \(BC=16\)
• \(m\angle A=64^{\circ} \), \(AB=29\), \(AC=34\)
• \(m\angle C=133^{\circ} \), \(m\angle B=25^{\circ} \),\(AB=48\)

Use the Law of Sines to solve \(\Delta ABC\) below.

• \(m\angle A=20^{\circ} \), \(AB=12\), \(BC=5\)

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.

• Let’s say we have \(\Delta ABC\) 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 \(m\angle A=20^{\circ} \), \(m\angle C=125^{\circ}\), \(AC=8.4\)
• Does the triangle that you found in problem 14 meet the requirements of the given information in problem 13? How are the two different \(m\angle C\) related? Draw the two possible triangles overlapping to visualize this relationship.

Review (Answers)

To view the Review answers, open this PDF file and look for section 8.10.

Additional Resources

Video: The Law of Sines: The Basics

Law of Sines Worksheet

Students will practice applying the law of sines to calculate side lengths and angle measurements. This worksheet includes word problems as well as challenging bonus problems.

Example Questions

Visual Aids

Other Details

This is a 5 part worksheet:

• Part I Model Problems
• Part II Practice Problems (1-6)
• Part III Practice (harder) & Word Problems (7 - 18)
• Part IV Challenge Problems
• Part V Answer Key
• Pictures of Law of Sines (triangles, formula and more..)
• Law of Sines and Cosines Worksheets

Ultimate Math Solver (Free) Free Algebra Solver ... type anything in there!

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Chapter 3: Laws of Sines and Cosines

Exercises: 3.2 The Law of Sines

• Use the law of sines to find a side #1-6
• Use the law of sines to find an angle #7-12
• Use the law of sines to solve an oblique triangle #13-18
• Solve problems using the law of sines #19-28
• Compute distances using the parallax #29-32
• Solve problems involving the ambiguous case #33-46

Suggested homework problems

Homework 3.2

Exercise group.

For Problems 13–18, sketch the triangle and solve. Round answers to two decimal places. 13. [latex]b = 7,~ A = 23°,~ B = 42°[/latex] 14. [latex]c = 34,~ A = 53°,~ C = 26°[/latex] 15. [latex]a = 1.8,~ c = 2.1,~ C = 44°[/latex] 16. [latex]b = 8.5,~ c = 6.8,~ B = 23°[/latex] 17. [latex]c = 75,~ A = 35°,~ B = 46°[/latex] 18. [latex]a = 94,~ B = 29°,~ C = 84°[/latex]

For Problems 19–26, sketch and label a triangle to illustrate the problem. Solve the problem. 19. Maryam wants to know the height of a cliff on the other side of a ravine. The angle of elevation from her edge of the ravine to the cliff top is [latex]84.6°{.}[/latex] When she moves [latex]30[/latex] feet back from the ravine, the angle of elevation is [latex]82.5°{.}[/latex] How tall is the cliff? 20. Amir wants to know the height of a tree in the median strip of a highway. The angle of elevation from the highway shoulder to the treetop is [latex]43.5°{.}[/latex] When he moves 10 feet farther away from the tree, the angle of elevation is [latex]37.2°{.}[/latex] How tall is the tree? 21. Delbert and Francine are [latex]10[/latex] kilometers apart, both observing a satellite that passes directly over their heads. At a moment when the satellite is between them, Francine measures its angle of elevation as [latex]84.6°{,}[/latex] and Delbert measures an angle of [latex]87°{.}[/latex] How far is the satellite from Delbert? 22. Megan rows her kayak due east. When she began, she spotted a lighthouse [latex]2000[/latex] meters in the distance at an angle of [latex]14°[/latex] south of east. After traveling for an hour, the lighthouse was at an angle of [latex]83°[/latex] south of east. How far did Megan travel, and what was her average speed? 23. Chad is hiking along a straight path but needs to detour around a large pond. He turns [latex]23°[/latex] from his path until clear of the pond, then walks back to his original path, intercepting it at an angle of [latex]29°[/latex] and at a distance of [latex]2[/latex] miles from where he had left the path. How far did Chad walk in each of the two segments of his detour, and how much farther did his detour require compared with a straight line through the pond?

• Find [latex]\angle ACB{.}[/latex]
• Find[latex]\angle CAB{,}[/latex] at the top of the antenna.
• How long is [latex]BC{,}[/latex] the distance from the bottom of the antenna to [latex]C{?}[/latex]
• How tall is the hill?

28. A billboard of California’s gubernatorial candidate Angelyne is located on the roof of a building. At a distance of [latex]180[/latex] feet from the building, the angles of elevation to the bottom and top of the billboard are, respectively, [latex]39.8°[/latex] and [latex]47.3°{.}[/latex] How tall is the billboard?

For Problems 29–32, compute the following distances in astronomical units (AUs). Then convert to kilometers, using the fact that 1 AU [latex]\approx 1.5[/latex] times [latex]10^{8}[/latex] km. 29. When observed from opposite sides of Earth’s orbit, the star Alpha Centauri has a parallax of [latex]0.76^{\prime\prime}{.}[/latex] How far from the Sun is Alpha Centauri? 30. How far from the Sun is Barnard’s Star, which has a parallax of [latex]1.1^{\prime\prime}[/latex] when observed at opposite ends of Earth’s orbit? 31. How far from the Sun is Tau Ceti, which has a parallax of [latex]0.55^{\prime\prime}[/latex] when observed from opposite ends of Earth’s orbit? 32. How far from the Sun is Sirius, which has a parallax of [latex]0.75^{\prime\prime}[/latex] when observed from opposite ends of Earth’s orbit?

• Use the definition of [latex]\sin A[/latex] to solve for [latex]a[/latex] (the length of side [latex]\overline{BC}[/latex]).
• Can you draw a triangle [latex]\triangle ABC[/latex] with [latex]\angle A = 30°[/latex] and [latex]c = 3[/latex] if [latex]a \lt \dfrac{3}{2}{?}[/latex] Why or why not?
• How many triangles are possible if [latex]\dfrac{3}{2} \lt a \lt 3{?}[/latex]
• How many triangles are possible if [latex]a \gt 3{?}[/latex]

34. In this problem we show that there are two different triangles [latex]\triangle ABC[/latex] with [latex]\angle A = 30°,~ a = 2[/latex] and [latex]c = 3{.}[/latex]

• Use a protractor to draw an angle [latex]\angle A = 30°{.}[/latex] Mark point [latex]B[/latex] on one side of the angle so that [latex]\overline{AB}[/latex] is 3 inches long.
• Locate two distinct points on the other side of the angle that are each 2 inches from point [latex]B{.}[/latex] These points are both possible locations for point [latex]C{.}[/latex]
• Use the law of sines to find two distinct possible measures for [latex]\angle C{.}[/latex]

35. In [latex]\triangle ABC, \angle A = 30°[/latex] and [latex]c = 12{.}[/latex] How many triangles are possible for each of the following lengths for side [latex]a{?}[/latex] Sketch the solutions in each case.

• [latex]a = 6[/latex]
• [latex]a = 4[/latex]
• [latex]a = 9[/latex]
• [latex]a = 15[/latex]
• Express the length of the altitude in terms of [latex]\angle A[/latex] and [latex]c{.}[/latex]
• Now suppose we keep [latex]\angle A[/latex] and side [latex]c[/latex] fixed but allow [latex]a[/latex] to vary in length. What is the smallest value [latex]a[/latex] can have and still be long enough to make a triangle?
• What are the largest and smallest values that [latex]a[/latex] can have in order to produce two distinct triangles [latex]\triangle ABC[/latex] (without changing [latex]\angle A[/latex] and side [latex]c[/latex])?

37. For the triangle in Problem 36, suppose [latex]\angle A = 40°[/latex] and [latex]c = 8{.}[/latex]

• Sketch and solve the triangle if [latex]a = 12{.}[/latex]
• Sketch and solve the triangle if [latex]a = 6{.}[/latex]
• Sketch and solve the triangle if [latex]a = 4{.}[/latex]
• For what value of [latex]a[/latex] is [latex]c[/latex] the hypotenuse of a right triangle?

38. For the figure in Problem 36, suppose [latex]\angle A = 70°[/latex] and [latex]c = 20{.}[/latex]

• For what value of [latex]a[/latex] is the triangle a right triangle?
• For what values of [latex]a[/latex] are there two solutions for the triangle?
• For what values of [latex]a[/latex] is there one obtuse solution for the triangle?
• For what value of [latex]a[/latex] is there no solution?

For Problems 39–42, find the remaining angles of the triangle. Round answers to two decimal places. (These problems involve the ambiguous case.) 39. [latex]a = 66,~ c = 43,~ \angle C = 25°[/latex] 40. [latex]b = 10,~ c = 14,~ \angle B = 20°[/latex] 41. [latex]b = 100,~ c = 80,~ \angle B = 49°[/latex] 42. [latex]b = 4.7,~ c = 6.3,~ \angle C = 54°[/latex] 43. Delbert and Francine are [latex]1000[/latex] yards apart. The angle Delbert sees between Francine and a certain tree is [latex]38°{.}[/latex] If the tree is [latex]800[/latex] yards from Francine, how far is it from Delbert? (There are two possible answers.) 44. From the lookout point on Fabrick Rock, Ann can see not only the famous “Crooked Spire” in Chesterfield, which is [latex]8[/latex] miles away, but also the red phone box in the village of Alton. Chesterfield and Alton are [latex]7[/latex] miles apart. Fabrick Rock has a plaque that shows directions to famous sites, and from the plaque Ann determines that the angle between the lines to the spire and the phone box measures [latex]19°{.}[/latex] How far is Fabrick Rock from the phone box? (There are two possible answers.) 45.

• Sketch a triangle with [latex]\angle A = 25°,~ \angle B = 35°{,}[/latex] and [latex]b = 16{.}[/latex]
• Use the law of sines to find [latex]a{.}[/latex]
• Use the law of sines to find [latex]c{.}[/latex]
• Find [latex]c[/latex] without using the law of sines. (Hint: Sketch the altitude, [latex]h{,}[/latex] from [latex]\angle C[/latex] to make two right triangles. Find [latex]h{,}[/latex] then use [latex]h[/latex] to find [latex]c[/latex].)
• Sketch a triangle with [latex]\angle A = 75°[/latex], [latex]a = 15{,}[/latex] and [latex]b = 6{.}[/latex]
• Find [latex]c[/latex] without using the law of sines.

Problems 47–48 prove the law of sines using the formula for the area of a triangle. (See Section 3.1 for the appropriate formula.) 47.

• Sketch a triangle with angles [latex]\angle A, ~\angle B[/latex] and [latex]\angle C[/latex] and opposite sides of lengths respectively [latex]a,~ b[/latex] and [latex]c{.}[/latex]
• Write the area of the triangle in terms of [latex]a,~b{,}[/latex] and angle [latex]\angle C{.}[/latex]
• Write the area of the triangle in terms of [latex]a,~c{,}[/latex] and angle [latex]\angle B{.}[/latex]
• Write the area of the triangle in terms of [latex]b

48. Equate the three different expressions from Problem 47 for the area of the triangle. Multiply through by [latex]\dfrac{2}{abc}[/latex] and simplify to deduce the law of sines. 49. Here is a method for solving certain oblique triangles by dividing them into two right triangles. In the triangle shown, we know two angles, [latex]\angle A[/latex] and [latex]\angle B{,}[/latex] and the side opposite one of them, say [latex]a{.}[/latex] We would like to find side [latex]b{.}[/latex]

• Draw the altitude [latex]h[/latex] from angle [latex]\angle C{.}[/latex]
• Write an expression for [latex]b[/latex] in terms of [latex]h[/latex] and angle [latex]\angle A{.}[/latex]
• Write an expression for [latex]h[/latex] in terms of angle [latex]\angle B{.}[/latex]
• Substitute your expression for [latex]h[/latex] into your expression for [latex]b{.}[/latex]
• [latex]a \sin A = b \sin B[/latex]
• [latex]\dfrac{a}{\sin A} = \dfrac{b}{\sin B}[/latex]
• [latex]\dfrac{a}{\sin B} = \dfrac{b}{\sin A}[/latex]

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Law of Sines Homework Help

Two residential buildings are to be constructed with a grassy recreational area between them. The taller building is 700 ft high. From the roof of the shorter building, the angle of elevation to the top of the taller building is 78° and the angle of depression to the base of the taller building is 62°.

• What is the height of the shorter building?
• What is the distance between the two buildings?

1 Expert Answer

Mark M. answered • 04/20/21

Mathematics Teacher - NCLB Highly Qualified

Draw and label a diagram!

d represents the distance between the two

h represents the height of the smaller

h represents the bottom of the taller opposite 62°

700 - h represents the top of the taller opposite 78°

tan 78° = (700 - h) / d

d tan 78 ° = 700 - h

d = (700 - h) / tan 78°

tan 62° = h / d

d tan 62 = h

d = h / 62°

Set the two equal and solve for h

Use the same tangent equations yet isolate h and solve for d.

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3.2: The Law of Cosines

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• Katherine Yoshiwara
• Los Angeles Pierce College

If we know two angles and one side of a triangle, we can use the Law of Sines to solve the triangle. We can also use the Law of Sines when we know two sides and the angle opposite one of them. But the Law of Sines is not helpful for the problem that opened this chapter, finding the distance from Avery to Clio. In this case we know two sides of the triangle, \(a\) and \(c\), and the included angle, \(B\).

To solve a triangle when we know two sides and the included angle, we will need a generalization of the Pythagorean theorem known as the Law of Cosines.

In a right triangle, with \(C = 90^{\circ}\), the Pythagorean theorem tells us that

\(c^2 = a^2 + b^2\)

If we allow angle \(C\) to vary, but keep \(a\) and \(b\) the same length, the side \(c\) will grow or shrink, depending on whether we increase or decrease the angle \(C\), as shown below.

The Pythagorean theorem is actually a special case of a more general law that applies to all triangles, no matter what the size of angle \(C\). The equation relating the three sides of a triangle is

\(c^2=a^2+b^2-2 a b \cos C\)

You can see that when \(C\) is a right angle, \(\cos 90^{\circ}=0\), so the equation reduces to the Pythagorean theorem.

We can write similar equations involving the angles or \(A\) or \(B\). The three equations are all versions of the Law of Cosines.

Law of Cosines.

If the angles of a triangle are \(A, B\), and \(C\), and the opposite sides are respectively \(a, b\), and \(c\), then

\begin{aligned} &a^2=b^2+c^2-2 b c \cos A \\ &b^2=a^2+c^2-2 a c \cos B \\ &c^2=a^2+b^2-2 a b \cos C \end{aligned}

Note 3.38 For a proof of the Law of Cosines, see Homework Problems 57 and 58.

Finding a Side

Now we can solve the problem of the distance from Avery to Clio. Here is the figure from Section 3.1 showing the location of the three towns.

Example 3.39

How far is it from Avery to Clio?

The angle \(\angle A B C=35^{\circ}+90^{\circ}=125^{\circ}\). Thus, in \(\triangle A B C\) we have \(a=34, c=48\) and \(B=125^{\circ}\). The distance from Avery to Clio is represented by \(b\) in the figure.

We know two sides and the included angle, and we choose the version of the Law of Cosines that uses our known angle, \(B\).

\begin{aligned} b^2 &= a^2 + c^2 - 2ac \cos B \quad \quad &&\text{Substitute the known values.} \\ b^2 &= 34^2 + 48^2 - 2(34)(48)\cos 125^{\circ} &&\text{Simplify the right side.} \\ b^2 &= 3460 - 3264\cos 125^{\circ} = 5332.153 &&\text{Take square roots.} \\ b &= 73.02 \end{aligned}

Avery is about 73 miles from Clio.

Caution 3.40

When simplifying the Law of Cosines, be careful to follow the order of operations. In the previous example, the right side of the equation

\(b^2=34^2+48^2-2(34)(48) \cos 125^{\circ}\)

has three terms, and simplifies to

\begin{aligned} &b^2=1156+2304-3264 \cos 125^{\circ} \\ &b^2=3460-3264(-0.573567364 \ldots) \end{aligned}

Note that 3264 is the coefficient of \(\cos 125^{\circ}\), so it would be incorrect to subtract 3264 from 3460. If you are using a graphing calculator, you can enter the right side of the equation exactly as it is written.

Checkpoint 3.41

In \(\triangle A B C, a=11, c=23\), and \(B=87^{\circ}\). Find \(b\), and round your answer to two decimal places.

Finding an Angle

We can also use the Law of Cosines to find an angle when we know all three sides of a triangle. Pay close attention to the algebraic steps used to solve the equation in the next example.

Example 3.42

In the triangle at right, \(a = 6, b = 7\), and \(c = 11\). Find angle \(C\).

We choose the version of the Law of Cosines that uses angle \(C\).

\begin{aligned} c^2 &= a^2 + b^2 - 2ab\cos C \quad \quad &&\text{Substitute the known values.} \\ 11^2 &= 6^2 + 7^2 - 2(6)(7)\cos C &&\text{Simplify each side.} \\ 121 &= 36 + 49 - 84\cos C &&\text{Isolate the cosine term.} \\ 36 &= -84\cos C &&\text{Solve for }\cos C. \\ \dfrac{-3}{7} &= \cos C &&\text{Solve for }C. \\ C &= \cos^{-1}\left(\dfrac{-3}{7}\right) = 115.4^{\circ} \end{aligned}

Angle \(C\) is about \(115.4^{\circ}\).

Checkpoint 3.43

In \(\triangle A B C, a=5.3, b=4.7\), and \(c=6.1\). Find angle \(B\), and round your answer to two decimal places.

\(48.07^{\circ}\)

Once we have calculated one of the angles in a triangle, we can use either the Law of Sines or the Law of Cosines to find a second angle. Here is how we would use the Law of Sines to find angle \(A\) in the previous example.

\begin{aligned} \dfrac{\sin A}{a} &=\dfrac{\sin C}{c} \quad \quad &&\text{Substitute the known values}.\\ \dfrac{\sin A}{6} &=\dfrac{\sin 115.4^{\circ}}{11} &&\text{Solve for }\sin A.\\ \sin A &=6 \cdot \dfrac{\sin 115.4^{\circ}}{11} \approx 0.4928 \end{aligned}

Thus, \(A=\sin ^{-1}(0.4928)=29.5^{\circ}\). (We know that \(A\) is an acute angle because it is opposite the shortest side of the triangle.) Finally,

\(B=180^{\circ}-(A+C) \approx 35.1^{\circ}\)

Alternatively, we can use the Law of Cosines to find angle \(A\).

\begin{aligned} a^2 &=b^2+c^2-2 b c \cos A & & \text { Substitute the known values. } \\ 6^2 &=7^2+11^2-2(7)(11) \cos A & & \text { Simplify each side. } \\ 36 &=49+121-154 \cos A & & \text { Isolate the cosine term. } \\ -134 &=-154 \cos A & & \text { Solve for } \cos \boldsymbol{A} . \\ \dfrac{67}{77} &=\cos A & & \text { Solve for A. } \\ A &=\cos ^{-1}\left(\dfrac{67}{77}\right)=29.5^{\circ} & & \end{aligned}

Note 3.44 Using the Law of Sines requires fewer calculations than the Law of Cosines, but the Law of Cosines uses only the original values, instead of the results of our previous calculations and approximations.

Whenever we round off a number, we introduce inaccuracy into the calculations, and these inaccuracies grow with each additional calculation. Thus, for the sake of accuracy, it is best to use given values in preference to calculated values whenever possible.

Using the Law of Cosines for the Ambiguous Case

In Section 3.2 we encountered the ambiguous case:

If we know two sides \(a\) and \(b\) of a triangle and the acute angle \(\alpha\) opposite one of them, there may be one solution, two solutions, or no solution, depending on the size of \(a\) in relation to \(b\) and \(\alpha\), as shown below.

The Ambiguous Case.

1 No solution: \(a < b \sin \alpha\)

\(\alpha\) is too short to make a triangle.

2 One solution: \(a = b \sin \alpha\)

\(\alpha\) is exactly the right length to make a right triangle.

3 Two solutions: \(b \sin \alpha < a < b\)

4 One solution: \(a > b\)

If \(\alpha\) is an obtuse angle, things are simpler: there is one solution if \(a>b\), and no solution if \(a \leq b\). Because there are always two angles with a given sine, if we use the Law of Sines for the ambiguous case, we must check whether both possible angles result in a triangle. But here is another approach: We can apply the Law of Cosines to find the third side first. With that method, we'll need the quadratic formula.

Quadratic Formula.

The solutions of the quadratic equation \(a x^2+b x+c=0, \quad a \neq 0\), are given by

\(x=\dfrac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

A quadratic equation can have one solution, two solutions, or no solution, depending on the value of the discriminant , \(b^2-4 a c\). If we use the Law of Cosines to find a side in the ambiguous case, the quadratic formula will tell us how many triangles have the given properties.

• If the quadratic equation has one positive solutions, there is one triangle.
• If the quadratic equation has two positive solutions, there are two triangles.
• If the quadratic equation has no positive solutions, there is no triangle with the given properties.

Example 3.45

In \(\triangle A B C, B=14.4^{\circ}, a=8\), and \(b=3\). Solve the triangle.

We begin by finding the third side of the triangle. Using the Law of Cosines, we have

\begin{aligned} b^2 &= a^2 + c^2 - 2ac \cos B \quad \quad &&\text{Substitute the known values.} \\ 3^2 &= 8^2 + c^2 - 2(8)c\cos 14.4^{\circ} &&\text{Simplify.} \\ 9 &= 64 + c^2 - 16c(0.9686) &&\text{Write the quadratic equation in standard from.} \\ 0 &= c^2 - 15.497c + 55 &&\text{Apply the quadratic formula.} \end{aligned}

\begin{aligned} c &= \dfrac{15.497 \pm \sqrt{(-15.497)^2 - 4(1)(55)}}{2(1)} \quad \quad &&\text{Simplify.} \\ &= \dfrac{15.497 \pm 4.490}{2} = 5.503 \text{ or } 9.994 \end{aligned}

Because there are two positive solutions for side \(c\), either \(c=5.503\) or \(c=9.994\), there are two triangles with the given properties. We apply the Law of Cosines again to find angle \(C\) in each triangle.

For the triangle with \(c=5.503\), we have

\begin{aligned} c^2 &= a^2 + b^2 - 2ab\cos C \quad \quad &&\text{Substitute the known values.} \\ 5.503^2 &= 8^2 +3^2 - 2(8)(3)\cos C &&\text{Solve for }\cos C. \\ \cos C &= \dfrac{5.503^2 - 64 - 9}{-48} = 0.889876 \\ C &= \cos^{-1} 0.889876) = 27.1^{\circ} \end{aligned}

and \(A = 180^{\circ} − (14.4^{\circ} + 27.1^{\circ} ) = 138.5^{\circ}\).

For the triangle with \(c = 9.994\), we have

\begin{aligned} c^2 &= a^2 + b^2 - 2ab\cos C \quad \quad &&\text{Substitute the known values.} \\ 9.994^2 &= 8^2 + 3^2 - 2(8)(3)\cos C &&\text{Solve for }\cos C. \\ \cos C &= \dfrac{9.994^2 - 64 - 9}{-48} = -0.560027 \\ C &= \cos^{-1}(-0.560027) = 124.1^{\circ} \end{aligned}

and \(A = 180^{\circ} - (14.4^{\circ} + 124.1^{\circ}) = 41.5^{\circ}\). Both triangles are shown below.

Note 3.46 In the previous example, notice that we used the Law of Cosines instead of the Law of Sines to find a second angle in the triangle, because there is only one angle between \(0^{\circ}\) and \(180^{\circ}\) with a given cosine. We don't have to check the results, as we would if we used the Law of Sines.

Checkpoint 3.47

Use the Law of Cosines to find all triangles \(A B C\) with \(A=48^{\circ}, a=10\), and \(b=15\).

No Solution.

Even with the aid of GPS (Global Positioning System) instruments, aircraft pilots and ship captains need to understand navigation based on trigonometry.

Example 3.48

The sailing club leaves the marina on a heading \(15^{\circ}\) east of north and sails for 18 miles. They then change course, and after traveling for 12 miles on a heading \(35^{\circ}\) east of north, they experience engine trouble and radio for help. The marina sends a speed boat to rescue them. How far should the speed boat go, and on what heading?

We'd like to find the distance \(x\) and the angle \(\theta\) shown in the figure. In \(\triangle A B C\), we can calculate the angle at point \(B\) where the sailing club changed course:

\(B = 180^{\circ} - 35^{\circ} + 15^{\circ} = 160^{\circ}\)

We know \(a = 12\) and \(c = 18\). We use the Law of Cosines to find \(b\) and \(\angle A\).

\begin{aligned} b^2 &= a^2 + c^2 - 2ac\cos B \quad \quad &&\text{Substitute the known values.} \\ &= 12^2 + 18^2 - 2(12)18\cos 160^{\circ} &&\text{Evaluate.} \\ &= 873.95 &&\text{Take positive square root.} \\ b&= 29.56 \end{aligned}

Next, we apply the Law of Cosines again to find \(\angle A\).

\begin{aligned} a^2 &=b^2+c^2-2 b c \cos A & & \text { Substitute the known values. } \\ 12^2 &=29.56^2+18^2-2(29.56)(18) \cos A & & \text { Solve for } \cos \boldsymbol{A} \\ \cos A &=\dfrac{12^2-29.56^2-18^2}{-2(29.56)(18)}=0.9903 & \\ C &=\cos ^{-1}(0.9903)=8^{\circ} & \end{aligned}

Thus, \(x=29.56\) and \(\theta=8^{\circ}+15^{\circ}=23^{\circ}\). The speed boat should travel \(29.56\) miles on a heading \(23^{\circ}\) east of north.

Checkpoint 3.49

Howard wants to fly from Anchorage to Nome, Alaska, a distance of 540 miles on a heading \(57^{\circ}\) west of north. After flying for some time, he discovers that his heading is in error, and he is actually flying \(47^{\circ}\) west of north. Howard corrects his flight plan and changes course when he is exactly 200 miles from Anchorage. What is his new heading, and how far is he from Nome?

\(62.8^{\circ}\) west of north, \(344.8\) miles

Which Law to Use

How can we decide which law, the Law of Sines or the Law of Cosines, is appropriate for a given problem?

• If we are solving a right triangle, we don’t need the Laws of Sines and Cosines; all we need are the definitions of the trigonometric ratios.
• But for oblique triangles, we can identify the following cases:

How to Solve an Oblique Triangle.

Note 3.50 In the ambiguous case, SSA, the Law of Sines is easier to apply, but there will be two possible angles, and we must check each angle to see if it produces a solution. Using the Law of Cosines involves solving a quadratic equation, but each positive solution of the equation yields a solution of the triangle.

Example 3.51

In the triangle at right, which law should you use to find \(\angle B\)? Which law should you use to find \(c\)?

We know two sides of the triangle and the angle opposite one of them. We can use the Law of Sines to find \(\angle B\).

\begin{aligned} \dfrac{\sin B}{6} &=\dfrac{\sin 40^{\circ}}{10} \\ \sin B &=0.3857 \end{aligned}

This is the ambiguous case; there are two angles between \(0^{\circ}\) and \(180^{\circ}\) with sine \(0.3857\), and we must check each angle to see if it produces a solution.

If instead we start by finding side \(c\), we use the Law of Cosines.

\begin{aligned} 10^2 &=6^2+c^2-2(6) c \cos 40^{\circ} \\ c^2-\left(12 \cos 40^{\circ}\right) c-64 &=0 \\ c &=\dfrac{12 \cos 40^{\circ} \pm \sqrt{\left(-12 \cos 40^{\circ}\right)^2-4(1)(-64)}}{2(1)} \end{aligned}

You can check that the quadratic equation has only one positive solution for \(c\) (or notice that because \(a>b\), angle \(B\) must be acute).

Checkpoint 3.52

In the triangle at right, which part of the triangle can you find, and which law should you use?

We first find side \(c\) using the Law of Cosines

Review the following skills you will need for this section.

Algebra Refresher 3.4

Solve each quadratic equation.

1 \(2.5x^2 + 6.2 = 816.2\)

2 \(0.8x^2 - 124 = 376\)

3 \(2x(2x-3) = 208\)

4 \(3x(x+5) = 900\)

5 \(2x^2 - 6x = 233.12\)

6 \(0.5x^2 + 1.5x = 464\)

1 \(\pm 8\)

2 \(\pm 25\)

3 \(8, -6.5\)

4 \(15, -20\)

5 \(12.4, -9.4\)

6 \(29, -32\)

Section 3.3 Summary

• Quadratic equation

• Quadratic formula

• Discriminant

1 The Law of Sines is not helpful when we know two sides of the triangle and the included angle. In this case we need the Law of Cosines.

2 If the angles of a triangle are \(A, B\), and \(C\), and the opposite sides are respectively \(a, b\), and \(c\), then

3 We can also use the Law of Cosines to find an angle when we know all three sides of a triangle.

4 We can use the Law of Cosines to solve the ambiguous case.

Study Questions

1 The Law of Cosines is really a generalization of what familiar theorem?

2 If you know all three sides of a triangle and one angle, what might be the advantage in using the Law of Cosines o find another angle, instead of the Law of Sines?

3 State the quadratic formula from memory. Try to sing the quadratic formula to the tune of ”Pop Goes the Weasel.”

4 Francine is solving a triangle in which \(a = 20, b = 16\), and \(A = 26^{\circ}\). She finds that \(\sin B = 0.3507\). How does she know that \(B = 20.5^{\circ}\), and not \(159.5^{\circ}\)?

1 Use the Law of Cosines to find the side opposite an angle #7-12

2 Use the Law of Cosines to find an angle #13-20

3 Use the Law of Cosines to find a side adjacent to an angle #21-26

4 Decide which law to use #27-34

5 Solve a triangle #35-42

6 Solve problems using the Law of Cosines #43-56

Homework 3.3

a Simplify \(5^2+7^2-2(5)(7) \cos \theta\) b Evaluate the expression in part (a) for \(\theta=29^{\circ}\) c Evaluate the expression in part (a) for \(\theta=151^{\circ}\)

a Simplify \(26.1^2+32.5^2-2(26.1)(32.5) \cos \phi\) b Evaluate the expression in part (a) for \(\phi=64^{\circ}\) c Evaluate the expression in part (a) for \(\phi=116^{\circ}\)

a Solve \(b^2=a^2+c^2-2 a c \cos \beta\) for \(\cos \beta\) b For the equation in part (a), find \(\cos \beta\) if \(a=5, b=11\), and \(c=8\).

a Solve \(a^2=b^2+c^2-2 b c \cos \alpha\) for \(\cos \alpha\) b For the equation in part (a), find \(\cos \alpha\) if \(a=4.6, b=7.2\), and \(c=9.4\).

a The equation \(9^2=b^2+4^2-2 b(4) \cos \alpha\) is quadratic in \(b\). Write the equation in standard form. b Solve the equation in part (a) for \(b\) if \(\alpha=48^{\circ}\).

a The equation \(5^2=6^2+c^2-2(5) c \cos \beta\) is quadratic in \(c\)

b Solve the equation in part (a) for \(c\) if \(\beta = 126^{\circ}.

For Problems 7–12, use the Law of Cosines to find the indicated side. Round to two decimal places.

For Problems 13–16,use the Law of Cosines to find the indicated angle. Round to two decimal places.

For Problems 17–20, find the angles of the triangle. Round answers to two decimal places.

17. \(a = 23, b = 14, c = 18\)

18. \(a = 18, b = 25, c = 19\)

19. \(a = 16.3, b = 28.1, c = 19.4\)

20. \(a = 82.3, b = 22.5, c = 66.8\)

For Problems 21–26, use the Law of Cosines to find the unknown side. Round your answers to two decimal places.

For Problems 27–34, which law should you use to find the labeled unknown value, the Law of Sines or the Law of Cosines? Write an equation you can solve to find the unknown value. For Problems 31-34, you may need two steps to find the unknown value.

For Problems 35–42,

a Sketch and label the triangle.

b Solve the triangle. Round answers to two decimal places.

35. \(B = 47^{\circ} , a = 23, c = 17\)

36. \(C = 32^{\circ} , a = 14, b = 18\)

37. \(a = 8, b = 7, c = 9\)

38. \(a = 23, b = 34, c = 45\)

39. \(b = 72, c = 98, B = 38^{\circ}\)

40. \(a = 28, c = 41, A = 27^{\circ}\)

41. \(c = 5.7, A = 59^{\circ} , B = 82^{\circ}\)

42. \(b = 82, A = 11^{\circ}, C = 42^{\circ}\)

a Sketch and label a triangle to illustrate the problem.

b Solve the problem. Round answers to one decimal place.

43. A surveyor would like to know the distance \(P Q\) across a small lake, as shown in the figure. She stands at point \(O\) and measures the angle between the lines of sight to points \(P\) and \(Q\) at \(76^{\circ}\). She also finds \(OP = 1400\) meters and \(OQ = 600\) meters. Calculate the distance \(P Q\).

44. Highway engineers plan to drill a tunnel through Boney Mountain from \(G\) to \(H\), as shown in the figure. The angle at point \(F\) is \(41^{\circ}\), and the distances to \(G\) and \(H\) are \(900\) yards and \(2500\) yards, respectively. How long will the tunnel be?

45. Two pilots leave an airport at the same time. One pilot flies \(3^{\circ}\) east of north at a speed of \(320\) miles per hour, the other flies \(157^{\circ}\) east of north at a speed of \(406\) miles per hour. How far apart are the two pilots after \(3\) hours? What is the heading from the first plane to the second plane at that time?

46. Two boats leave port at the same time. One boat sails due west at a speed of \(17\) miles per hour, the other powers \(42^{\circ}\) east of north at a speed of \(23\) miles per hour. How far apart are the two boats after \(2\) hours? What is the heading from the first boat to the second boat at that time?

46. Two boats leave port at the same time. One boat sails due west at a speed of 17 miles per hour, the other powers \(42^{\circ}\) east of north at a speed of 23 miles per hour. How far apart are the two boats after 2 hours? What is the heading from the first boat to the second boat at that time?

47. Caroline wants to fly directly south from Indianapolis to Cancun, Mexico, a distance of 1290 miles. However, to avoid bad weather, she flies for 400 miles on a heading \(18^{\circ}\) east of south. What is the heading to Cancun from that location, and how far is it?

48. Alex sails 8 miles from Key West, Florida on a heading \(40^{\circ}\) east of south. He then changes course and sails for 10 miles due east. What is the heading back to Key West from that point, and how far is it?

49. The phone company wants to erect a cell tower on a steep hill inclined \(26^{\circ}\) to the horizontal. The installation crew plans to run a guy wire from a point on the ground 20 feet uphill from the base of the tower and attach it to the tower at a height of 100 feet. How long should the guy wire be?

50. Sandstone Peak rises 3500 above the desert. The Park Service plans to run an aerial tramway up the north face, which is inclined at an angle of \(68^{\circ}\) to the horizontal. The base station will be located 500 feet from the foot of Sandstone Peak. Ignoring any slack in the cable, how long should it be?

51. The sides of a triangle are 27 cm, 15 cm, and 20 cm. Find the area of the triangle. (Hint: Find one of the angles first.)

52. The sides of a parallelogram are 10 inches and 8 inches, and form an angle of \(130^{\circ}\). Find the lengths of the diagonals of the parallelogram.

For Problems 53–56, find x, the distance from one vertex to the foot of the altitude.

Problems 57 and 58 prove the Law of Cosines.

a Copy the three figures above showing the three possibilities for an angle \(C\) in a triangle: \(C\) is acute, obtuse, or a right angle. For each figure, explain why it is true that \(c^2=(b-x)^2+y^2\), then rewrite the right side to get \(c^2=\left(x^2+y^2\right)+b^2-2 b x\).

b For each figure, explain why it is true that \(x^2+y^2=a^2\).

c For all three figures, \(a\) is the distance from the origin to the point \((x, y)\). Use the definition of cosine to write \(\cos C\) in terms of \(a\) and \(x\), then solve your equation for \(x\).

d Start with the last equation from (a), and substitute expressions from (b) and (c) to conclude one case of the Law of Cosines.

58. Demonstrate the other two cases of the Law of Cosines:

• \(a^2=b^2+c^2-2 b c \cos A\)
• \(b^2=a^2+c^2-2 a c \cos B\)

(Hint: See Problem 57 and switch the roles of \(a\) and \(c\), etc.)

59. Use the Law of Cosines to prove the projection laws:

\begin{aligned} a &=b \cos C+c \cos B \\ b &=c \cos A+a \cos C \\ c &=a \cos B+b \cos A \end{aligned}

Illustrate with a sketch. (Hint: Add together two of the versions of the Law of Cosines.)

60. If \(\triangle A B C\) is isosceles with \(a=b\), show that \(c^2=2 a^2(1-\cos C)\).

61. Use the Law of Cosines to prove:

\begin{aligned} 1+\cos A &=\dfrac{(a+b+c)(-a+b+C)}{2 b c} \\ 1-\cos A &=\dfrac{(a-b+c)(-a+b+C)}{2 b c} \end{aligned}

62. Prove that

\(\dfrac{\cos A}{a}+\dfrac{\cos B}{b}+\dfrac{\cos C}{c}=\dfrac{a^2+b^2+c^2}{2 a b c}\)