law of sines assignment quizlet

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Precalculus

Course: precalculus   >   unit 2.

• Trig word problem: stars
• General triangle word problems

Laws of sines and cosines review

Law of sines

Law of cosines, practice set 1: solving triangles using the law of sines, example 1: finding a missing side, example 2: finding a missing angle.

• Your answer should be
• an integer, like 6 ‍  
• a simplified proper fraction, like 3 / 5 ‍  
• a simplified improper fraction, like 7 / 4 ‍  
• a mixed number, like 1   3 / 4 ‍  
• an exact decimal, like 0.75 ‍  
• a multiple of pi, like 12   pi ‍   or 2 / 3   pi ‍  

Practice set 2: Solving triangles using the law of cosines

Example 1: finding an angle, example 2: finding a missing side, practice set 3: general triangle word problems, want to join the conversation.

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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.

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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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

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Determine the number of possible triangles that can be drawn with the following dimensions: m<B = 73, a = 7 and b = 5.

For which scenario would you need to use the ambiguous case?

Determine the number of possible triangles that can be drawn with the following dimensions: m<B = 33, a=27, and b = 22.

Determine the number of possible triangles that can be drawn with the following dimensions: m<B = 29, a = 14 and b = 19.

Which of the following scenarios would you be unable to use Law of Sines?

In ΔABC, m<C = 53, m<B = 44 and AC = 7. Find AB to nearest tenth.

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Module 13: Non-Right Triangles

Introduction to law of sines, learning objectives.

By the end of this section, you will be able to:

• Use the Law of Sines to solve oblique triangles.
• Solve applied problems using the Law of Sines.

Suppose two radar stations located 20 miles apart each detect an aircraft between them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles .

• Precalculus. Authored by : OpenStax College. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected]:1/Preface . License : CC BY: Attribution

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5.2: The Law of Sines

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• Carl Stitz & Jeff Zeager
• Lakeland Community College & Lorain County Community College

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Trigonometry literally means ‘measuring triangles’ and with Chapter 10 under our belts, we are more than prepared to do just that. The main goal of this section and the next is to develop theorems which allow us to ‘solve’ triangles – that is, find the length of each side of a triangle and the measure of each of its angles. In Sections 10.2 , 10.3 and 10.6 , we’ve had some experience solving right triangles. The following example reviews what we know.

Example 11.2.1

Given a right triangle with a hypotenuse of length 7 units and one leg of length 4 units, find the length of the remaining side and the measures of the remaining angles. Express the angles in decimal degrees, rounded to the nearest hundreth of a degree.

For definitiveness, we label the triangle below.

To find the length of the missing side \(a\), we use the Pythagorean Theorem to get \(a^{2}+4^{2}=7^{2}\) which then yields \(a=\sqrt{33}\) units. Now that all three sides of the triangle are known, there are several ways we can find \(\alpha\) using the inverse trigonometric functions. To decrease the chances of propagating error, however, we stick to using the data given to us in the problem. In this case, the lengths 4 and 7 were given, so we want to relate these to \(\alpha\). According to Theorem 10.4 , \(\cos (\alpha)=\frac{4}{7}\). Since \(\alpha\) is an acute angle, \(\alpha=\arccos \left(\frac{4}{7}\right)\) radians. Converting to degrees, we find \(\alpha \approx 55.15^{\circ}\). Now that we have the measure of angle \(\alpha\), we could find the measure of angle \(\beta\) using the fact that \(\alpha\) and \(\beta\) are complements so \(\alpha+\beta=90^{\circ}\). Once again, we opt to use the data given to us in the problem. According to Theorem 10.4 , we have that \(\sin (\beta)=\frac{4}{7}\) so \(\beta=\arcsin \left(\frac{4}{7}\right)\) radians and we have \(\beta \approx 34.85^{\circ}\).

A few remarks about Example 11.2.1 are in order. First, we adhere to the convention that a lower case Greek letter denotes an angle 1 and the corresponding lowercase English letter represents the side 2 opposite that angle. Thus, \(a\) is the side opposite \(\alpha, b\) is the side opposite \(\beta\) and \(c\) is the side opposite \(\gamma\). Taken together, the pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\) are called angle-side opposite pairs. Second, as mentioned earlier, we will strive to solve for quantities using the original data given in the problem whenever possible. While this is not always the easiest or fastest way to proceed, it minimizes the chances of propagated error. 3 Third, since many of the applications which require solving triangles ‘in the wild’ rely on degree measure, we shall adopt this convention for the time being. 4 The Pythagorean Theorem along with Theorems 10.4 and 10.10 allow us to easily handle any given right triangle problem, but what if the triangle isn’t a right triangle? In certain cases, we can use the Law of Sines to help.

Theorem 11.2. The Law of Sines

Given a triangle with angle-side opposite pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\), the following ratios hold

\(\frac{\sin (\alpha)}{a}=\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\)

or, equivalently,

\(\frac{a}{\sin (\alpha)}=\frac{b}{\sin (\beta)}=\frac{c}{\sin (\gamma)}\)

The proof of the Law of Sines can be broken into three cases. For our first case, consider the triangle \(\triangle A B C\) below, all of whose angles are acute, with angle-side opposite pairs \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\). If we drop an altitude from vertex \(B\), we divide the triangle into two right triangles: \(\triangle A B Q\) and \(\triangle B C Q\). If we call the length of the altitude \(h\) (for height), we get from Theorem 10.4 that \(\sin (\alpha)=\frac{h}{c}\) and \(\sin (\gamma)=\frac{h}{a}\) so that \(h=c \sin (\alpha)=a \sin (\gamma)\). After some rearrangement of the last equation, we get \(\frac{\sin (\alpha)}{a}=\frac{\sin (\gamma)}{c}\). If we drop an altitude from vertex \(A\), we can proceed as above using the triangles \(\triangle A B Q\) and \(\triangle A C Q\) to get \(\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\), completing the proof for this case.

For our next case consider the triangle \(\triangle A B C\) below with obtuse angle \(\alpha\). Extending an altitude from vertex \(A\) gives two right triangles, as in the previous case: \(\triangle A B Q\) and \(\triangle A C Q\). Proceeding as before, we get \(h=b \sin (\gamma)\) and \(h=c \sin (\beta)\) so that \(\frac{\sin (\beta)}{b}=\frac{\sin (\gamma)}{c}\).

Dropping an altitude from vertex B also generates two right triangles, \(\triangle A B Q\) and \(\triangle B C Q\). We know that \(\sin \left(\alpha^{\prime}\right)=\frac{h^{\prime}}{c}\) so that \(h^{\prime}=c \sin \left(\alpha^{\prime}\right)\). Since \(\alpha^{\prime}=180^{\circ}-\alpha, \sin \left(\alpha^{\prime}\right)=\sin (\alpha)\), so in fact, we have \(h^{\prime}=c \sin (\alpha)\). Proceeding to \(\triangle B C Q\), we get \(\sin (\gamma)=\frac{h^{\prime}}{a} \text { so } h^{\prime}=a \sin (\gamma)\). Putting this together with the previous equation, we get \(\frac{\sin (\gamma)}{c}=\frac{\sin (\alpha)}{a}\), and we are finished with this case.

The remaining case is when \(\triangle A B C\) is a right triangle. In this case, the Law of Sines reduces to the formulas given in Theorem 10.4 and is left to the reader. In order to use the Law of Sines to solve a triangle, we need at least one angle-side opposite pair. The next example showcases some of the power, and the pitfalls, of the Law of Sines.

Example 11.2.2

Solve the following triangles. Give exact answers and decimal approximations (rounded to hundredths) and sketch the triangle.

• \(\alpha=120^{\circ}, a=7 \text { units, } \beta=45^{\circ}\)
• \(\alpha=85^{\circ}, \beta=30^{\circ}, c=5.25 \text { units }\)
• \(\alpha=30^{\circ}, a=1 \text { units, } c=4 \text { units }\)
• \(\alpha=30^{\circ}, a=2 \text { units, } c=4 \text { units }\)
• \(\alpha=30^{\circ}, a=3 \text { units, } c=4 \text { units }\)
• \(\alpha=30^{\circ}, a=4 \text { units, } c=4 \text { units }\)
• Knowing an angle-side opposite pair, namely \(\alpha\) and \(a\), we may proceed in using the Law of Sines. Since \(\beta=45^{\circ}\), we use \(\frac{b}{\sin \left(45^{\circ}\right)}=\frac{7}{\sin \left(120^{\circ}\right)}\) so \(b=\frac{7 \sin \left(45^{\circ}\right)}{\sin \left(120^{\circ}\right)}=\frac{7 \sqrt{6}}{3} \approx 5.72\) units. Now that we have two angle-side pairs, it is time to find the third. To find \(\gamma\), we use the fact that the sum of the measures of the angles in a triangle is \(180^{\circ}\). Hence, \(\gamma=180^{\circ}-120^{\circ}-45^{\circ}=15^{\circ}\). To find \(c\), we have no choice but to used the derived value \(\gamma=15^{\circ}\), yet we can minimize the propagation of error here by using the given angle-side opposite pair \((\alpha, a)\). The Law of Sines gives us \(\frac{c}{\sin \left(15^{\circ}\right)}=\frac{7}{\sin \left(120^{\circ}\right)}\) so that \(c=\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)} \approx 2.09\) units. 5

• Since we are given \((\alpha, a)\) and \(c\), we use the Law of Sines to find the measure of \(\gamma\). We start with \(\frac{\sin (\gamma)}{4}=\frac{\sin \left(30^{\circ}\right)}{1}\) and get \(\sin (\gamma)=4 \sin \left(30^{\circ}\right)=2\). Since the range of the sine function is [−1, 1], there is no real number with \(\sin (\gamma)=2\). Geometrically, we see that side \(a\) is just too short to make a triangle. The next three examples keep the same values for the measure of \(\alpha\) and the length of \(c\) while varying the length of \(a\). We will discuss this case in more detail after we see what happens in those examples.

Some remarks about Example 11.2.2 are in order. We first note that if we are given the measures of two of the angles in a triangle, say \(\alpha\) and \(\beta\), the measure of the third angle \(\gamma\) uniquely determined using the equation \(\gamma=180^{\circ}-\alpha-\beta\). Knowing the measures of all three angles of a triangle completely determines its shape. If in addition we are given the length of one of the sides of the triangle, we can then use the Law of Sines to find the lengths of the remaining two sides to determine the size of the triangle. Such is the case in numbers 1 and 2 above. In number 1, the given side is adjacent to just one of the angles – this is called the ‘Angle-Angle-Side’ (AAS) case. 8 In number 2, the given side is adjacent to both angles which means we are in the so-called ‘Angle-Side-Angle’ (ASA) case. If, on the other hand, we are given the measure of just one of the angles in the triangle along with the length of two sides, only one of which is adjacent to the given angle, we are in the ‘Angle-Side-Side’ (ASS) case. 9 In number 3, the length of the one given side \(a\) was too short to even form a triangle; in number 4, the length of a was just long enough to form a right triangle; in 5, \(a\) was long enough, but not too long, so that two triangles were possible; and in number 6, side \(a\) was long enough to form a triangle but too long to swing back and form two. These four cases exemplify all of the possibilities in the Angle-Side-Side case which are summarized in the following theorem.

Theorem 11.3

Suppose \((\alpha, a)\) and \((\gamma, c)\) are intended to be angle-side pairs in a triangle where \(\alpha\), \(a\) and \(c\) are given. Let \(h=c \sin (\alpha)\).

• If \(a<h\), then no triangle exists which satisfies the given criteria.
• If \(a=h\), then \(\gamma=90^{\circ}\) so exactly one (right) triangle exists which satisfies the criteria.
• If \(h<a<c\), then two distinct triangles exist which satisfy the given criteria.
• If \(a \geq c\), then \(\gamma\) is acute and exactly one triangle exists which satisfies the given criteria

Theorem 11.3 is proved on a case-by-case basis. If \(a<h\), then \(a<c \sin (\alpha)\). If a triangle were to exist, the Law of Sines would have \(\frac{\sin (\gamma)}{c}=\frac{\sin (\alpha)}{a}\) so that \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}>\frac{a}{a}=1\), which is impossible. In the figure below, we see geometrically why this is the case.

Simply put, if \(a<h\) the side \(a\) is too short to connect to form a triangle. This means if \(a \geq h\), we are always guaranteed to have at least one triangle, and the remaining parts of the theorem tell us what kind and how many triangles to expect in each case. If \(a = h\), then \(a=c \sin (\alpha)\) and the Law of Sines gives \(\frac{\sin (\alpha)}{a}=\frac{\sin (\gamma)}{c}\) so that \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}=\frac{a}{a}=1\). Here, \(\gamma=90^{\circ}\) as required. Moving along, now suppose \(h<a<c\). As before, the Law of Sines 10 gives \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}\). Since \(h<a\), \(c \sin (\alpha)<a\) or \(\frac{c \sin (\alpha)}{a}<1\) which means there are two solutions to \(\sin (\gamma)=\frac{c \sin (\alpha)}{a}\): an acute angle which we’ll call \(\gamma_{0}\), and its supplement, \(180^{\circ}-\gamma_{0}\). We need to argue that each of these angles ‘fit’ into a triangle with \(\alpha\). Since \((\alpha, a)\) and \(\left(\gamma_{0}, c\right)\) are angle-side opposite pairs, the assumption \(c>a\) a in this case gives us \(\gamma_{0}>\alpha\). Since \(\gamma_{0}\) is acute, we must have that \(\alpha\) is acute as well. This means one triangle can contain both \(\alpha\) and \(\gamma_{0}\), giving us one of the triangles promised in the theorem. If we manipulate the inequality \(\gamma_{0}>\alpha\) a bit, we have \(180^{\circ}-\gamma_{0}<180^{\circ}-\alpha\) which gives \(\left(180^{\circ}-\gamma_{0}\right)+\alpha<180^{\circ}\). This proves a triangle can contain both of the angles \(\alpha\) and \(\left(180^{\circ}-\gamma_{0}\right)\), giving us the second triangle predicted in the theorem. To prove the last case in the theorem, we assume \(a \geq c\). Then \(\alpha \geq \gamma\), which forces \(\gamma\) to be an acute angle. Hence, we get only one triangle in this case, completing the proof.

Example 11.2.3

Sasquatch Island lies off the coast of Ippizuti Lake. Two sightings, taken 5 miles apart, are made to the island. The angle between the shore and the island at the first observation point is \(30^{\circ}\) and at the second point the angle is \(45^{\circ}\). Assuming a straight coastline, find the distance from the second observation point to the island. What point on the shore is closest to the island? How far is the island from this point?

We sketch the problem below with the first observation point labeled as \(P\) and the second as \(Q\). In order to use the Law of Sines to find the distance \(d\) from \(Q\) to the island, we first need to find the measure of \(\beta\) which is the angle opposite the side of length 5 miles. To that end, we note that the angles \(\gamma\) and \(45^{\circ}\) are supplemental, so that \(\gamma=180^{\circ}-45^{\circ}=135^{\circ}\). We can now find \(\beta=180^{\circ}-30^{\circ}-\gamma=180^{\circ}-30^{\circ}-135^{\circ}=15^{\circ}\). By the Law of Sines, we have \(\frac{d}{\sin \left(30^{\circ}\right)}=\frac{5}{\sin \left(15^{\circ}\right)}\) which gives \(d=\frac{5 \sin \left(30^{\circ}\right)}{\sin \left(15^{\circ}\right)} \approx 9.66\) miles. Next, to find the point on the coast closest to the island, which we’ve labeled as \(C\), we need to find the perpendicular distance from the island to the coast. 11

Let \(x\) denote the distance from the second observation point \(Q\) to the point \(C\) and let \(y\) denote the distance from \(C\) to the island. Using Theorem 10.4 , we get \(\sin \left(45^{\circ}\right)=\frac{y}{d}\). After some rearranging, we find \(y=d \sin \left(45^{\circ}\right) \approx 9.66\left(\frac{\sqrt{2}}{2}\right) \approx 6.83\) miles. Hence, the island is approximately 6.83 miles from the coast. To find the distance from \(Q\) to \(C\), we note that \(\beta=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}\) so by symmetry, 12 we get \(x=y \approx 6.83\) miles. Hence, the point on the shore closest to the island is approximately 6.83 miles down the coast from the second observation point.

We close this section with a new formula to compute the area enclosed by a triangle. Its proof uses the same cases and diagrams as the proof of the Law of Sines and is left as an exercise.

Theorem 11.4

Suppose \((\alpha, a),(\beta, b) \text { and }(\gamma, c)\) are the angle-side opposite pairs of a triangle. Then the area \(A\) enclosed by the triangle is given by

\(A=\frac{1}{2} b c \sin (\alpha)=\frac{1}{2} a c \sin (\beta)=\frac{1}{2} a b \sin (\gamma)\)

Example 11.2.4

Find the area of the triangle in Example 11.2.2 number 1.

From our work in Example 11.2.2 number 1, we have all three angles and all three sides to work with. However, to minimize propagated error, we choose \(A=\frac{1}{2} a c \sin (\beta)\) from Theorem 11.4 because it uses the most pieces of given information. We are given \(a=7\) and \(\beta=45^{\circ}\), and we calculated \(c=\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\). Using these values, we find \(A=\frac{1}{2}(7)\left(\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\right) \sin \left(45^{\circ}\right)=\approx 5.18\) square units. The reader is encouraged to check this answer against the results obtained using the other formulas in Theorem 11.4 .

11.2.1 Exercises

In Exercises 1 - 20, solve for the remaining side(s) and angle(s) if possible. As in the text, \((\alpha, a)\), \((\beta, b)\) and \((\gamma, c)\) are angle-side opposite pairs.

• \(\alpha=13^{\circ}, \beta=17^{\circ}, a=5\)
• \(\alpha=73.2^{\circ}, \beta=54.1^{\circ}, a=117\)
• \(\alpha=95^{\circ}, \beta=85^{\circ}, a=33.33\)
• \(\alpha=95^{\circ}, \beta=62^{\circ}, a=33.33\)
• \(\alpha=117^{\circ}, a=35, b=42\)
• \(\alpha=117^{\circ}, a=45, b=42\)
• \(\alpha=68.7^{\circ}, a=88, b=92\)
• \(\alpha=42^{\circ}, a=17, b=23.5\)
• \(\alpha=68.7^{\circ}, a=70, b=90\)
• \(\alpha=30^{\circ}, a=7, b=14\)
• \(\alpha=42^{\circ}, a=39, b=23.5\)
• \(\gamma=53^{\circ}, \alpha=53^{\circ}, c=28.01\)
• \(\alpha=6^{\circ}, a=57, b=100\)
• \(\gamma=74.6^{\circ}, c=3, a=3.05\)
• \(\beta=102^{\circ}, b=16.75, c=13\)
• \(\beta=102^{\circ}, b=16.75, c=18\)
• \(\beta=102^{\circ}, \gamma=35^{\circ}, b=16.75\)
• \(\beta=29.13^{\circ}, \gamma=83.95^{\circ}, b=314.15\)
• \(\gamma=120^{\circ}, \beta=61^{\circ}, c=4\)
• \(\alpha=50^{\circ}, a=25, b=12.5\)
• Find the area of the triangles given in Exercises 1, 12 and 20 above.

(Another Classic Application: Grade of a Road) The grade of a road is much like the pitch of a roof (See Example 10.6.6 ) in that it expresses the ratio of rise/run. In the case of a road, this ratio is always positive because it is measured going uphill and it is usually given as a percentage. For example, a road which rises 7 feet for every 100 feet of (horizontal) forward progress is said to have a 7% grade. However, if we want to apply any Trigonometry to a story problem involving roads going uphill or downhill, we need to view the grade as an angle with respect to the horizontal. In Exercises 22 - 24, we first have you change road grades into angles and then use the Law of Sines in an application.

• Using a right triangle with a horizontal leg of length 100 and vertical leg with length 7, show that a \(7 \%\) grade means that the road (hypotenuse) makes about a \(4^{\circ}\) angle with the horizontal. (It will not be exactly \(4^{\circ}\), but it’s pretty close.)
• What grade is given by a \(9.65^{\circ}\) angle made by the road and the horizontal? 13
• Along a long, straight stretch of mountain road with a 7% grade, you see a tall tree standing perfectly plumb alongside the road. 14 From a point 500 feet downhill from the tree, the angle of inclination from the road to the top of the tree is \(6^{\circ}\). Use the Law of Sines to find the height of the tree. (Hint: First show that the tree makes a \(94^{\circ}\) angle with the road.)

(Another Classic Application: Bearings) In the next several exercises we introduce and work with the navigation tool known as bearings. Simply put, a bearing is the direction you are heading according to a compass. The classic nomenclature for bearings, however, is not given as an angle in standard position, so we must first understand the notation. A bearing is given as an acute angle of rotation (to the east or to the west) away from the north-south (up and down) line of a compass rose. For example, \(\mathrm{N} 40^{\circ} \mathrm{E}\) (read "\(40^{\circ}\) east of north”) is a bearing which is rotated clockwise \(40^{\circ}\) from due north. If we imagine standing at the origin in the Cartesian Plane, this bearing would have us heading into Quadrant I along the terminal side of \(\theta=50^{\circ}\). Similarly, \(\mathrm{S} 50^{\circ} \mathrm{W}\) would point into Quadrant III along the terminal side of \(\theta=220^{\circ}\) because we started out pointing due south (along \(\theta=270^{\circ}\)) and rotated clockwise \(50^{\circ}\) back to \(220^{\circ}\). Counter-clockwise rotations would be found in the bearings \(\mathrm{N} 60^{\circ} \mathrm{W}\) (which is on the terminal side of \(\theta=150^{\circ}\)) and \(\mathrm{S} 27^{\circ} \mathrm{E}\) (which lies along the terminal side of \(\theta=297^{\circ}\)). These four bearings are drawn in the plane below.

The cardinal directions north, south, east and west are usually not given as bearings in the fashion described above, but rather, one just refers to them as ‘due north’, ‘due south’, ‘due east’ and ‘due west’, respectively, and it is assumed that you know which quadrantal angle goes with each cardinal direction. (Hint: Look at the diagram above.)

• \(\mathrm{S} 83^{\circ} \mathrm{E}\)
• \(\mathrm{N} 5.5^{\circ} \mathrm{E}\)
• \(\mathrm{N} 31.25^{\circ} \mathrm{W}\)
• \(\mathrm{S}{72}^{\circ} 41^{\prime} 12^{\prime \prime} \mathrm{W}\) 15
• \(\mathrm{N} 45^{\circ} \mathrm{E}\)
• \(\mathrm{S}{45}{ }^{\circ} \mathrm{W}\)
• The Colonel spots a campfire at a of bearing \(\mathrm{N} 42^{\circ} \mathrm{E}\) from his current position. Sarge, who is positioned 3000 feet due east of the Colonel, reckons the bearing to the fire to be \(\mathrm{N} 20^{\circ} \mathrm{W}\) from his current position. Determine the distance from the campfire to each man, rounded to the nearest foot.
• A hiker starts walking due west from Sasquatch Point and gets to the Chupacabra Trailhead before she realizes that she hasn’t reset her pedometer. From the Chupacabra Trailhead she hikes for 5 miles along a bearing of \(\mathrm{N} 53^{\circ} \mathrm{W}\) which brings her to the Muffin Ridge Observatory. From there, she knows a bearing of \(\mathrm{S}6 5^{\circ} \mathrm{E}\) will take her straight back to Sasquatch Point. How far will she have to walk to get from the Muffin Ridge Observatory to Sasquach Point? What is the distance between Sasquatch Point and the Chupacabra Trailhead?
• The captain of the SS Bigfoot sees a signal flare at a bearing of \(\mathrm{N} 15^{\circ} \mathrm{E}\) from her current location. From his position, the captain of the HMS Sasquatch finds the signal flare to be at a bearing of \(\mathrm{N} 75^{\circ} \mathrm{W}\). If the SS Bigfoot is 5 miles from the HMS Sasquatch and the bearing from the SS Bigfoot to the HMS Sasquatch is \(\mathrm{N} 50^{\circ} \mathrm{E}\), find the distances from the flare to each vessel, rounded to the nearest tenth of a mile.
• Carl spies a potential Sasquatch nest at a bearing of \(\mathrm{N} 10^{\circ} \mathrm{E}\) and radios Jeff, who is at a bearing of \(\mathrm{N} 50^{\circ} \mathrm{E}\) from Carl’s position. From Jeff’s position, the nest is at a bearing of \(\mathrm{S} 70^{\circ} \mathrm{W}\). If Jeff and Carl are 500 feet apart, how far is Jeff from the Sasquatch nest? Round your answer to the nearest foot.
• A hiker determines the bearing to a lodge from her current position is \(\mathrm{S} 40^{\circ} \mathrm{W}\). She proceeds to hike 2 miles at a bearing of \(\mathrm{S} 20^{\circ} \mathrm{E}\) at which point she determines the bearing to the lodge is \(\mathrm{S}{75}{ }^{\circ} \mathrm{W}\). How far is she from the lodge at this point? Round your answer to the nearest hundredth of a mile.
• A watchtower spots a ship off shore at a bearing of \(\mathrm{N} 70^{\circ} \mathrm{E}\). A second tower, which is 50 miles from the first at a bearing of \(\mathrm{S} 80^{\circ} \mathrm{E}\) from the first tower, determines the bearing to the ship to be \(\mathrm{N} 25^{\circ} \mathrm{W}\). How far is the boat from the second tower? Round your answer to the nearest tenth of a mile.
• Skippy and Sally decide to hunt UFOs. One night, they position themselves 2 miles apart on an abandoned stretch of desert runway. An hour into their investigation, Skippy spies a UFO hovering over a spot on the runway directly between him and Sally. He records the angle of inclination from the ground to the craft to be \(75^{\circ}\) and radios Sally immediately to find the angle of inclination from her position to the craft is \(50^{\circ}\). How high off the ground is the UFO at this point? Round your answer to the nearest foot. (Recall: 1 mile is 5280 feet.)
• The angle of depression from an observer in an apartment complex to a gargoyle on the building next door is \(55^{\circ}\). From a point five stories below the original observer, the angle of inclination to the gargoyle is \(20^{\circ}\). Find the distance from each observer to the gargoyle and the distance from the gargoyle to the apartment complex. Round your answers to the nearest foot. (Use the rule of thumb that one story of a building is 9 feet.)
• Prove that the Law of Sines holds when \(\triangle A B C\) is a right triangle.
• Discuss with your classmates why knowing only the three angles of a triangle is not enough to determine any of the sides.
• Discuss with your classmates why the Law of Sines cannot be used to find the angles in the triangle when only the three sides are given. Also discuss what happens if only two sides and the angle between them are given. (Said another way, explain why the Law of Sines cannot be used in the SSS and SAS cases.)
• the information yields no triangle
• the information yields exactly one right triangle
• the information yields two distinct triangles
• the information yields exactly one obtuse triangle

Explain why you cannot choose \(a\) in such a way as to have \(\alpha=30^{\circ}, b=10\) and your choice of \(a\) yield only one triangle where that unique triangle has three acute angles.

• Use the cases and diagrams in the proof of the Law of Sines ( Theorem 11.2 ) to prove the area formulas given in Theorem 11.4 . Why do those formulas yield square units when four quantities are being multiplied together?

11.2.2 Answers

• \(\begin{array}{lll} \alpha=13^{\circ} & \beta=17^{\circ} & \gamma=150^{\circ} \\ a=5 & b \approx 6.50 & c \approx 11.11 \end{array}\)
• \(\begin{array}{lll} \alpha=73.2^{\circ} & \beta=54.1^{\circ} & \gamma=52.7^{\circ} \\ a=117 & b \approx 99.00 & c \approx 97.22 \end{array}\)
• Information does not produce a triangle
• \( \begin{array}{lll} \alpha=95^{\circ} & \beta=62^{\circ} & \gamma=23^{\circ} \\ a=33.33 & b \approx 29.54 & c \approx 13.07 \end{array}\)
• \(\begin{array}{lll} \alpha=117^{\circ} & \beta \approx 56.3^{\circ} & \gamma \approx 6.7^{\circ} \\ a=45 & b=42 & c \approx 5.89 \end{array}\)
• \(\begin{array}{lll} \alpha=68.7^{\circ} & \beta \approx 76.9^{\circ} & \gamma \approx 34.4^{\circ} \\ a=88 & b=92 & c \approx 53.36 \end{array}\)
• \(\begin{array}{lll} \alpha=42^{\circ} & \beta \approx 67.66^{\circ} & \gamma \approx 70.34^{\circ} \\ a=17 & b=23.5 & c \approx 23.93 \end{array}\)
• \(\begin{array}{lll} \alpha=30^{\circ} & \beta=90^{\circ} & \gamma=60^{\circ} \\ a=7 & b=14 & c=7 \sqrt{3} \end{array}\)
• \(\begin{array}{lll} \alpha=42^{\circ} & \beta \approx 23.78^{\circ} & \gamma \approx 114.22^{\circ} \\ a=39 & b=23.5 & c \approx 53.15 \end{array}\)
• \(\begin{array}{lll} \alpha=53^{\circ} & \beta=74^{\circ} & \gamma=53^{\circ} \\ a=28.01 & b \approx 33.71 & c=28.01 \end{array}\)
• \(\begin{array}{lll} \alpha=6^{\circ} & \beta \approx 169.43^{\circ} & \gamma \approx 4.57^{\circ} \\ a=57 & b=100 & c \approx 43.45 \end{array}\)
• \(\begin{array}{lll} \alpha \approx 78.59^{\circ} & \beta \approx 26.81^{\circ} & \gamma=74.6^{\circ} \\ a=3.05 & b \approx 1.40 & c=3 \end{array}\)
• \(\begin{array}{lll} \alpha \approx 28.61^{\circ} & \beta=102^{\circ} & \gamma \approx 49.39^{\circ} \\ a \approx 8.20 & b=16.75 & c=13 \end{array}\)
• \(\begin{array}{lll} \alpha=43^{\circ} & \beta=102^{\circ} & \gamma=35^{\circ} \\ a \approx 11.68 & b=16.75 & c \approx 9.82 \end{array}\)
• \(\begin{array}{lll} \alpha=66.92^{\circ} & \beta=29.13^{\circ} & \gamma=83.95^{\circ} \\ a \approx 593.69 & b=314.15 & c \approx 641.75 \end{array}\)
• \(\begin{array}{lll} \alpha=50^{\circ} & \beta \approx 22.52^{\circ} & \gamma \approx 107.48^{\circ} \\ a=25 & b=12.5 & c \approx 31.13 \end{array}\)

The area of the triangle from Exercise 12 is about 377.1 square units.

The area of the triangle from Exercise 20 is about 149 square units.

• \(\arctan \left(\frac{7}{100}\right) \approx 0.699 \text { radians }\), which is equivalent to \(\(4.004^{\circ}\)
• About 53 feet
• \(\theta=180^{\circ}\)
• \(\theta=353^{\circ}\)
• \(\theta=84.5^{\circ}\)
• \(\theta=270^{\circ}\)
• \(\theta=121.25^{\circ}\)
• \(\theta=197^{\circ} 18^{\prime} 48^{\prime \prime}\)
• \(\theta=45^{\circ}\)
• \(\theta=225^{\circ}\)

Sarge is about 2525 feet to the campfire.

The distance from Sasquatch Point to the Chupacabra Trailhead is about 2.46 miles.

The HMS Sasquatch is about 2.9 miles from the flare.

• Jeff is about 371 feet from the nest.
• She is about 3.02 miles from the lodge
• The boat is about 25.1 miles from the second tower.
• The UFO is hovering about 9539 feet above the ground.

The gargoyle is about 27 feet from the observer on the lower floor.

The gargoyle is about 25 feet from the other building.

1 as well as the measure of said angle

2 as well as the length of said side

3 Your Science teachers should thank us for this.

4 Don’t worry! Radians will be back before you know it!

5 The exact value of \(\sin \left(15^{\circ}\right)\) could be found using the difference identity for sine or a half-angle formula, but that becomes unnecessarily messy for the discussion at hand. Thus “exact” here means \(\frac{7 \sin \left(15^{\circ}\right)}{\sin \left(120^{\circ}\right)}\).

6 To find an exact expression for \(\beta\), we convert everything back to radians: \(\alpha=30^{\circ}=\frac{\pi}{6}\) radians, \(\gamma=\arcsin \left(\frac{2}{3}\right)\) radians and \(180^{\circ}=\pi\) radians. Hence, \(\beta=\pi-\frac{\pi}{6}-\arcsin \left(\frac{2}{3}\right)=\frac{5 \pi}{6}-\arcsin \left(\frac{2}{3}\right) \text { radians } \approx 108.19^{\circ}\).

7 An exact answer for \(\beta\) in this case is \(\beta=\arcsin \left(\frac{2}{3}\right)-\frac{\pi}{6} \text { radians } \approx 11.81^{\circ}\).

8 If this sounds familiar, it should. From high school Geometry, we know there are four congruence conditions for triangles: Angle-Angle-Side (AAS), Angle-Side-Angle (ASA), Side-Angle-Side (SAS) and Side-Side-Side (SSS). If we are given information about a triangle that meets one of these four criteria, then we are guaranteed that exactly one triangle exists which satisfies the given criteria.

9 In more reputable books, this is called the ‘Side-Side-Angle’ or SSA case.

10 Remember, we have already argued that a triangle exists in this case!

11 Do you see why \(C\) must lie to the right of \(Q\)?

12 Or by Theorem 10.4 again . . .

13 I have friends who live in Pacifica, CA and their road is actually this steep. It’s not a nice road to drive.

14 The word ‘plumb’ here means that the tree is perpendicular to the horizontal.

15 See Example 10.1.1 in Section 10.1 for a review of the DMS system.

Law of Sines and Cosines

How to determine which formula to use, when to use law of sines vs cosines.

Practice Problems

Law of Sines

Just look at it . You can always immediately look at a triangle and tell whether or not you can use the Law of Sines.

You need either 2 sides and the non-included angle or, in this case, 2 angles and the non-included side.

The law of sines is all about opposite pairs.

In this case, we have a side of length 11 opposite a known angle of $$ 29^{\circ} $$ (first opposite pair) and we want to find the side opposite the known angle of $$ 118^\circ$$.

First Step $ \frac{\red x} {sin(118^{\circ})} = \frac{11}{ sin(29^{\circ})} $

Law of Cosines

Remember, the law of cosines is all about included angle (or knowing 3 sides and wanting to find an angle).

In this case, we have a side of length 20 and of 13 and the included angle of $$ 66^\circ$$.

You need either 2 sides and the non-included angle (like this triangle) or 2 angles and the non-included side.

Remember, the law of sines is all about opposite pairs.

In this case, we have a side of length 16 opposite a known angle of $$ 115^{\circ} $$ (first opposite pair) and we want to find the angle opposite the known side of length 32 . We can set up the proportion below and solve :

Decide which formula (Law of Sines/Cosines) you would use to calculate the value of $$ \red x$$ below? After you decide that, try to set up the equation (Do not solve -- just substitute into the proper formula).

Decide which formula (Law of Sines/Cosines) you would use to calculate the value of $$ \red x $$ below? After you decide that, try to set up the equation (Do not solve -- just substitute into the proper formula).

First Step $ \red x^2 = 11^2 + 7^2 -2(11)(7) \cdot cos(50) $

Decide which formula (Law of Sines/Cosines) you would use to calculate the value of x below? After you decide that, try to set up the equation (Do not solve -- just substitute into the proper formula).

First Step $ \frac{sin ( \red x)} {7 } = \frac{sin(50)}{11} $

• Worksheet on law of sines and law of cosines (pdf)

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Linear Algebra, Law of Sines and Cosines

Law of sines and cosines.