Law of Sines

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M4 PRECALCULUS AND ADVANCED TOPICS

Lesson 8: Law of Sines

Student Outcomes . Students prove the law of sines and use it to solve problems (G-SRT.D.10).

Lesson Notes In previous lessons, students developed tools for finding a missing side or a missing angle in a right triangle. In particular, students know how to use the Pythagorean theorem to find a missing side of a right triangle, and they know how to use the trigonometric functions to find a missing angle in a right triangle. In this lesson, students extend their knowledge to cover certain oblique triangles. Students develop a proof of the law of sines and apply it to solve problems. At the end of the lesson, students develop a second proof of the law of sines that uses properties of circles. In the course of the lesson, students draw on their knowledge of the conditions that imply a unique triangle, which students know as the triangle congruence criteria (ASA, SSS, SAS). This topic was addressed in Geometry Module 1 Lessons 22, 24, and 25.

Classwork Exercises 1–2 (2 minutes) In these exercises, students apply their prior knowledge of right triangle trigonometry to solve problems.

Exercises

1. Find the value of 풙 in the figure on the left.

ퟖ.ퟒ 퐬퐢퐧(ퟑퟑ°) = 풙 ퟖ. ퟒ 풙 = ≈ ퟏퟓ. ퟒ 퐬퐢퐧(ퟑퟑ°)

2. Find the value of 휶 in the figure on the right. ퟒ.ퟐ 퐬퐢퐧(휶) = ퟗ.ퟏ 휶 ≈ ퟐퟕ°

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This work is licensed under a This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. This file derived from PreCal-M4-TE-1.3.0 -10.2015

NYS COMMON CORE MATHEMATICS CURRICULUM Lesson 8 M4 PRECALCULUS AND ADVANCED TOPICS

Exploratory Challenge (9 minutes): Oblique Triangles . In Exercises 1 and 2, what aspects of the triangle were you given, and what aspects did you find?  In Exercise 1, we were given a side and an angle in a right triangle. From these measurements, we determined the length of the hypotenuse.  In Exercise 2, we were given two sides in a right triangle. From these measurements, we determined the measure of one of the acute angles.

8.4

41° C 72° B . Now examine the triangle above. What information is provided in this triangle?  We are given two angles and a non-included side. . How is this triangle different from the ones presented in the Exercises 1 and 2?  The problems in Exercises 1 and 2 involved right triangles, but this triangle does not have a right angle. . Do you think you should be able to determine the remaining measurements in this triangle? Why or why not? Think about this for a moment, then explain your reasoning to a neighbor.  180° MP.3 The missing angle can be found easily using the fact that the sum of the angles is .  The AAS criterion for triangle congruence tells us that only one triangle can be formed with the given information so the missing side measurements can be determined by the measurements that are given in the figure. . See how many of the missing measurements you can determine. Take several minutes to explore this problem MP.1 with the students around you. Give students several minutes to work on this problem in groups of 3 or 4. Select one or more students to present their findings to the class.

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Give students the opportunity to draw an altitude like the one shown below without being Scaffolding: prompted to do so. If students do not see that an auxiliary line is necessary, use the cues . If students need support in the scaffolding box on the right. with this task, prompt them with one or both of A the following questions. The goal is to get students to realize that they can make progress by drawing in one of the altitudes, 8.4 h which allows them to capitalize on their 41° knowledge of right C triangles. 72° . B What kind of auxiliary line could you draw here, and  When we draw the altitude to side 퐵퐶̅̅̅̅, we get two right triangles as where could you draw it? shown above. . It would be useful to be ℎ Looking at the yellow triangle, we get sin(72°) = . This means that able to draw on our 8.4 knowledge of right ℎ = 8.4 sin(72°). triangles. How might you ℎ Looking at the green triangle, we get sin(41°) = . This means that create right triangles in 퐴퐶 ℎ = 퐴퐶 ∙ sin(41°). this figure? Thus we have 퐴퐶 ∙ sin(41°) = 8.4 sin(72°), which means that 8.4sin(72°) 퐴퐶 = ≈ 12.2. sin(41°)  To make 180°, the third angle must be 67°. When we draw in the altitude to side ̅퐵퐴̅̅̅, we get a different pair of right triangles as shown below.

67°

12.2 8.4 k

C 72° B

푘 Looking at the blue triangle, we get sin(67°) = . This means that 푘 = 12.2 sin(67°). 12.2 푘 Looking at the red triangle, we get sin(72°) = . This means that 푘 = 퐵퐶 ∙ sin(72°). 퐵퐶 12.2sin(67°) Thus, we have 퐵퐶 ∙ sin(72°) = 12.2 sin(67°), which means that 퐵퐶 = ≈ 11.8. sin(72°)

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 Here is a picture of the triangle with the three original measurements, together with the three measurements we calculated above.

67°

12.2 8.4

41° C 72° 11.8 B . To summarize, we can use what we know about right triangles to learn things about oblique triangles, that is, triangles that do not have a right angle.

Exercise 3 (5 minutes) Instruct students to solve the following problem and then to compare their work with a partner. Select two students to write their work on the board to share with the class.

3. Find all of the measurements for the triangle below.

4.2

40° B 78° C The measure of ∠푨 is ퟔퟐ°. The length of side 푩푨̅̅̅̅ is ퟔ. ퟒ. The length of side 푩푪̅̅̅̅ is ퟓ. ퟖ.

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Discussion (5 minutes): Proving the Law of Sines . Can you generalize your findings? Try to find an equation that shows the relationship between the quantities in the figure below. Give students several minutes to work together. Select a student to share his or her work with the class.

a b

β α

 First we draw an altitude:

a b h

β α

One expression for the altitude is ℎ = 푏 ∙ sin⁡(훼). Another expression for the altitude is ℎ = 푎 ∙ sin⁡(훽). From these observations, it follows that 푎 ∙ sin(훽) = 푏 ∙ sin⁡(훼). We can use this equation to find 푎 if we know the value of 푏 and vice versa. . The fact that 푎 ∙ sin(훽) = 푏 ∙ sin⁡(훼) is called the law of sines. Since the sine function is a central feature of the formula, this name makes sense. Let’s be explicit about the meaning of the symbols in this equation. How are these quantities related in the figure?  The side labeled 푎 is opposite the angle labeled 훼, and the side labeled 푏 is opposite the angle labeled 훽. . By way of summary, let’s make the reasoning absolutely clear. Why are the expressions 푎 ∙ sin(훽) and 푏 ∙ sin⁡(훼) equal to each other? MP.2  Each of these expressions represents the length of an altitude in the triangle. Since they represent the same thing, the expressions must be equal to each other. . Now let’s get some practice applying the law of sines. Keep in mind that you can always draw an altitude just like you did above, but perhaps it is also worth knowing that the equation 푎 ∙ sin(훽) = 푏 ∙ sin⁡(훼) gives us a shortcut. 푎 푏 . When applying the law of sines, it is often easiest to work with the equation = . Can you see sin(훼) sin(훽) that this is equivalent to 푎 ∙ sin(훽) = 푏 ∙ sin⁡(훼)? Explain why these are equivalent.  By taking the equation 푎 ∙ sin(훽) = 푏 ∙ sin⁡(훼) and dividing both sides by sin(훼) ∙ sin⁡(훽), you get 푎 푏 = . sin(훼) sin(훽)

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Exercises 4–5 (4 minutes) Instruct students to solve the following problems and then to compare their work with a partner. Select two students to write their work on the board to share with the class.

4. Find the length of side 푨푪̅̅̅̅ in the triangle below.

C 55° B 32°

7.5

푨푪 ퟕ.ퟓ ퟕ.ퟓ⁡∙⁡퐬퐢퐧(ퟑퟐ°) We have⁡ = , which means 푨푪 = ≈ ퟒ. ퟗ. 퐬퐢퐧(ퟑퟐ°) 퐬퐢퐧(ퟓퟓ°) 퐬퐢퐧(ퟓퟓ°)

5. A hiker at point 푪 is ퟕ. ퟓ⁡kilometers from a hiker at point 푩; a third hiker is at point 푨. Use the angles shown in the diagram below to determine the distance between the hikers at points 푪 and 푨.

B 7.5 60°

C 34°

In order to apply the law of sines, first we need to find the third angle. The measure of the missing angle is ퟖퟔ°. 푨푪 ퟕ.ퟓ ퟕ.ퟓ⁡∙⁡퐬퐢퐧(ퟔퟎ°) Now we can use the law of sines, which gives⁡ = , which means 푨푪 = ≈ ퟔ. ퟓ. Thus, the 퐬퐢퐧(ퟔퟎ°) 퐬퐢퐧(ퟖퟔ°) 퐬퐢퐧(ퟖퟔ°) two hikers at points 푪 and 푨 are about ퟔ. ퟓ⁡퐤퐦 apart.

Discussion (7 minutes): Finding Angles

11.5

32°

6.3

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. Examine the triangle on the previous page. What information is given to you in this triangle? How is this situation different from the ones we encountered earlier in the lesson?  We are given two sides and a non-included angle. In the previous examples, we were given two angles and a side. . Do you think you should be able to determine the remaining measurements in this triangle? Why or why not? Think about this for a moment, then explain your reasoning to a neighbor. MP.3  This is the SSA case. We have to allow for the possibility that there are two triangles with these specific measurements. . Let’s have a look at the two triangles with these measurements:

11.5 11.5

32° 32° 6.3 6.3 α2

α1

. Notice that both triangles have a side with length 11.5, an angle with measure 32°, and a side with length 6.3 that is opposite the 32° angle. . As you can see in the figure, the angle across from the side with length 11.5 can be either acute or obtuse.

What can we say about the relationship between 훼1 and 훼2? Take a moment to think about this, and then discuss what you notice with a partner.  The figure contains an isosceles triangle, so its base angles must be congruent.

11.5

32° 6.3 6.3 α2 α1

α1

 Now we can see that 훼1 + 훼2 = 180. . Let’s apply the law of sines and then see if we can determine how to get both of these angles. 11.5 6.3 11.5∙⁡sin(32°)  We have⁡ = , which gives sin(훼) = ≈ 0.97. sin(훼) sin(32°) 6.3

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. Okay, so now we want to find values of 훼 for which sin(훼) ≈ 0.97. To see that there are indeed two such values, let’s look at a diagram of the unit circle:

. Evidently there are two ways we can rotate the point (1, 0) so that its image has a height equal to that of the dotted line.

. Can we see that 훼1 + 훼2 = 180 in this diagram as well?

 Yes, the two angles labeled 훼1 are congruent by a reflection across the 푦-axis. Thus, 훼1 + 훼2 = 180. 11.5⁡∙⁡sin(32°) . Okay, now let’s use a calculator to produce the two values for which sin(훼) = . 6.3

 We find that 훼1 ≈ 75.3°, so 훼2 = 180 − 75.3 = 104.7. 훼2 = 104.7°

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. Let’s return now to the original problem: What are the two possible values of 훼 in the figure below? Draw the two triangles with the measurements shown in the figure.

11.5

32°

6.3

 We found that 훼 could be 75.3°, which would give an acute triangle as shown on the left, but 훼 could also be 104.7°, which would give the obtuse triangle shown on the right.

11.5 11.5

32° 32° 6.3

6.3 104.7°

75.3°

Exercises 6–7 (3 minutes) Instruct students to solve the following problems and then to compare their work with a partner. Select two students to write their work on the board to share with the class.

6. Two sides of a triangle have lengths ퟏퟎ. ퟒ and ퟔ. ퟒ. The angle opposite ퟔ. ퟒ is ퟑퟔ°. What could the angle opposite ퟏퟎ. ퟒ be? ퟏퟎ.ퟒ ퟔ.ퟒ Let 휷 represent the angle opposite ퟏퟎ. ퟒ. Then, we have = , which gives 퐬퐢퐧⁡휷 ≈ ퟎ. ퟗퟓퟓ and 퐬퐢퐧(휷) 퐬퐢퐧(ퟑퟔ°) 휷 = ퟕퟐ. ퟖ°. We could also have 휷 = ퟏퟎퟕ. ퟐ° because an angle and its supplement have the same sine value.

7. Two sides of a triangle have lengths ퟗ. ퟔ and ퟏퟏ. ퟏ. The angle opposite ퟗ. ퟔ is ퟓퟗ°. What could the angle opposite ퟏퟏ. ퟏ be? ퟏퟏ.ퟏ ퟗ.ퟔ Let 휷 represent the angle opposite ퟏퟏ. ퟏ. Then, we have = , which gives 휷 = ퟖퟐ. ퟑ°. We could also 퐬퐢퐧(휷) 퐬퐢퐧(ퟓퟗ°) have 휷 = ퟗퟕ. ퟕ°.

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Discussion (Optional): Circumscribed Circles . One of the real joys of mathematics is that we can look at a problem through different lenses. Sometimes we develop new insights by using a different perspective, and this can be exciting! Let’s return to the problem we solved earlier in the lesson.

8.4

41° C 72° B . Recall that we were able to calculate the length of side 퐴퐶̅̅̅̅ by drawing the Scaffolding: altitude to side 퐵퐶̅̅̅̅. Another way of finding the missing side lengths is to draw . Students may need to be the circle that circumscribes the triangle: reminded of Thales’ theorem, which says that A an angle inscribed in a semicircle must be a right angle. . Students may need to be 8.4 reminded that an angle inscribed in a circle has 41° C half the measure of its 72° intercepted arc. B

. Let’s try to use what we know about circles to find 퐴퐶̅̅̅̅. The key here is to draw in the diameters one at a time. . Remember, the inscribed angle that includes the diameter must be a right angle, so use that knowledge to create the diameters.

A A

D D

8.4 8.4

41° C C

B B

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. Point 퐷 is the opposite end of the diameter through point 퐵. What observations can you make about △ 퐴퐵퐷? Take a minute to think about this. In particular, can you find the angles in △ 퐴퐵퐷?  Since ̅퐵퐷̅̅̅ is a diameter of the circle, it follows that △ 퐴퐵퐷 has a right angle at 퐴. We can also infer that ∠퐷 is 41° because it intercepts the same arc as ∠퐶. . So we’ve distorted △ 퐴퐵퐷 in such a way that it becomes a right triangle, but the distortion preserves the length of ̅퐴퐵̅̅̅ and the measure of ∠퐶. Clever, isn’t it? Now we can use right triangle trigonometry to describe the length of 퐵퐷̅̅̅̅. Can you see how to do this? 8.4  We have . This means 퐵퐷 = . sin(41°)

A A

C C 72° B B 72°

. Now let’s run a diameter through point 퐴 and conduct a similar analysis.  Using the same reasoning as before, we conclude that △ 퐴퐶퐸 is a right triangle and that ∠퐸 is equal to 퐴퐶 ∠퐵. Now we can describe the length of diameter⁡퐴퐸̅̅̅̅ using the sine function: sin(72°) = . This 퐴퐸⁡ 퐴퐶 gives us 퐴퐸 = . sin(72°) 8.4 . Now let’s put it all together: We know that diameter has length 퐵퐷 = and that diameter 퐴퐸̅̅̅̅ satisfies sin(41°) 퐴퐶 퐴퐸 = . How can we use these observations to compute the length of side 퐴퐶̅̅̅̅? Take a minute to really sin(72°) think about this!  The length of the diameter of a circle is constant, so we have 퐵퐷 = 퐴퐸. This means that 8.4 퐴퐶 8.4⁡∙⁡sin(72°) ⁡ = ⁡, and so 퐴퐶 = ≈ 12.2. sin(41°) sin(72°) sin(41°) . Is this consistent with the result we found at the start of the lesson?  Yes, we got 12.2 in that case also.

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Discussion (3 minutes): The Law of Sines via Circles . Let’s generalize the method involving circles.

a a

푎 푎 . This pair of diagrams shows that the diameter, 푑, of the circle satisfies sin(훼) = . This gives 푑 = . 푑 sin(훼) . What does this pair of diagrams show? 푏 푏  These diagrams show that the diameter, 푑, of the circle satisfies sin(훽) = . This gives 푑 = . 푑 sin(훽) . And what do we get when we combine the two observations about the diameters? 푎 푏  Since the diameter of the circle is constant, we have = . sin(훼) sin(훽) 푐 . In the same way, we can show that the diameter of the circle is described by the expression . sin(훾)

b a

β c

. Thus, we can express the law of sines in this form: If sides 푎, 푏, and 푐 are opposite angles 훼, 훽, and 훾, 푎 푏 푐 respectively, then the measurements satisfy the equation = = . sin(훼) sin(훽) sin(훾)

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sin(훼) sin(훽) sin(훾) . Could we also write = = ? Why or why not? 푎 푏 푐  Yes. If three fractions are equal, then their reciprocals are also equal. . It is worth knowing that it is sometimes convenient to set up a problem using this alternate form.

Closing (3 minutes) . State the law of sines, and explain its uses. 푎 푏 푐  The law of sines says that⁡ = = . sin(훼) sin(훽) sin(훾)  We can use this law to find the measurements of a triangle when we know two angles and one of the opposite sides or when we know two sides and one of the opposite angles. . If 푏 and 푐 are sides of a triangle that are opposite angles 훽 and 훾, we can write 푏 ∙ sin(훾) = 푐 ∙ sin(훽). What do these two expressions represent geometrically?  Each of these expressions represents the length of an altitude of the triangle. Since the expressions represent the same segment, they must be equal to each other. 푏 푐 . If 푏 and 푐 are sides of a triangle that are opposite angles 훽 and 훾, we can write⁡ = . What do these sin(훽) sin(훾) two expressions represent geometrically?  Each of these expressions represents the diameter of the circle that circumscribes the triangle. Since the diameter of a circle is constant, the expressions must be equal to each other.

Exit Ticket (4 minutes)

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Name Date

Lesson 8: Law of Sines

Exit Ticket

1. Find the length of side 퐴퐶̅̅̅̅ in the triangle below.

41°

C 105° 11.7 B

2. A triangle has sides with lengths 12.6 and 7.9. The angle opposite 7.9 is 37°. What are the possible values of the measure of the angle opposite 12.6?

7.9 7.9 12.6

? ? 37°

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Exit Ticket Sample Solutions

1. Find the length of side 푨푪̅̅̅̅ in the triangle below.

41°

C 105° 11.7 B

푨푪 ퟏퟏ.ퟕ ퟏퟏ.ퟕ⁡∙⁡퐬퐢퐧(ퟏퟎퟓ°) We have⁡ = , which gives 푪 = ≈ ퟏퟕ. ퟐ . 퐬퐢퐧(ퟏퟎퟓ°) 퐬퐢퐧(ퟒퟏ°) 퐬퐢퐧(ퟒퟏ°)

2. A triangle has sides with lengths ퟏퟐ. ퟔ and ퟕ. ퟗ. The angle opposite ퟕ. ퟗ is ퟑퟕ°. What are the possible values of the measure of the angle opposite ퟏퟐ. ퟔ?

7.9 7.9 12.6

? ? 37°

퐬퐢퐧(풙) 퐬퐢퐧(ퟑퟕ°) ퟏퟐ.ퟔ⁡∙⁡퐬퐢퐧(ퟑퟕ°) We have = , which means that 퐬퐢퐧(풙) = . The values of 풙 that satisfy this equation ퟏퟐ.ퟔ ퟕ.ퟗ ퟕ.ퟗ are 풙 ≈ ퟕퟑ. ퟕ° and 풙 ≈ ퟏퟎퟔ. ퟑ°.

Problem Set Sample Solutions

1. Let △ 푨푩푪 be a triangle with the given lengths and angle measurements. Find all possible missing measurements using the law of sines. a. 풂 = ퟓ, 풎∠푨 = ퟒퟑ°, 풎∠푩 = ퟖퟎ°.

풃 ≈ ퟕ. ퟐퟐ, 풄 ≈ ퟔ. ퟏퟓ, 풎∠푪 = ퟓퟕ°

b. 풂 = ퟑ. ퟐ, 풎∠푨 = ퟏퟏퟎ°, 풎∠푩 = ퟑퟓ°.

풃 ≈ ퟏ. ퟗퟓ, 풄 ≈ ퟏ. ퟗퟓ, 풎∠푪 = ퟑퟓ°

c. 풂 = ퟗ. ퟏ, 풎∠푨 = ퟕퟎ°, 풎∠푩 = ퟗퟓ°.

풃 ≈ ퟗ. ퟔퟓ, 풄 ≈ ퟐ. ퟓퟏ, 풎∠푪 = ퟏퟓ°

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d. 풂 = ퟑ. ퟐ, 풎∠푩 = ퟑퟎ°, 풎∠푪 = ퟒퟓ°.

풎∠푨 = ퟏퟎퟓ°, 풃 ≈ ퟏ. ퟔퟔ, 풄 ≈ ퟐ. ퟑퟒ

e. 풂 = ퟏퟐ, 풎∠푩 = ퟐퟗ°, 풎∠푪 = ퟑퟏ°.

풎∠푨 = ퟏퟐퟎ°, 풃 ≈ ퟔ. ퟕퟐ, 풄 ≈ ퟕ. ퟏퟒ

f. 풂 = ퟒ. ퟕ, 풎∠푩 = ퟏퟖ. ퟖ°, 풎∠푪 = ퟕퟐ°.

풎∠푨 = ퟖퟗ. ퟐ°, 풃 ≈ ퟏ. ퟓퟏ, 풄 ≈ ퟒ. ퟒퟕ

g. 풂 = ퟔ, 풃 = ퟑ, 풎∠푨 = ퟗퟏ°.

풎∠푩 ≈ ퟐퟗ. ퟗퟗ°, 풎∠푪 ≈ ퟓퟗ. ퟎퟏ°, 풄 ≈ ퟓ. ퟏퟒ

h. 풂 = ퟕ. ퟏ, 풃 = ퟕ, 풎∠푨 = ퟕퟎ°.

풎∠푩 = ퟔퟕ. ퟖퟗ°, 풎∠푪 ≈ ퟒퟐ. ퟏퟏ°, 풄 = ퟓ. ퟎퟕ

i. 풂 = ퟖ, 풃 = ퟓ, 풎∠푨 = ퟒퟓ°.

풎∠푩 = ퟐퟔ. ퟐퟑ°, 풎∠푪 = ퟏퟎퟖ. ퟕퟕ°, 풄 = ퟏퟎ. ퟕퟏ

j. 풂 = ퟑ. ퟓ, 풃 = ퟑ. ퟔ, 풎∠푨 = ퟑퟕ°.

풎∠푩 = ퟑퟖ. ퟐퟒ°, 풎∠푪 = ퟏퟎퟒ. ퟕퟔ°, 풄 = ퟓ. ퟔퟐ or 풎∠푩 = ퟏퟒퟏ. ퟕퟔ°, 풎∠푪 = ퟏ. ퟐퟒ°, 풄 = ퟎ. ퟏퟑ

k. 풂 = ퟗ, 풃 = ퟏퟎ. ퟏ, 풎∠푨 = ퟔퟏ°.

풎∠푩 = ퟕퟖ. ퟗퟕ°, 풎∠푪 = ퟒퟎ. ퟎퟑ°, 풄 = ퟔ. ퟔퟐ or 풎∠푩 = ퟏퟎퟏ. ퟎퟑ°, 풎∠푪 = ퟏퟕ. ퟗퟕ°, 풄 = ퟑ. ퟏퟕ

l. 풂 = ퟔ, 풃 = ퟖ, 풎∠푨 = ퟒퟏ. ퟓ°.

풎∠푩 = ퟔퟐ. ퟎퟕ°, 풎∠푪 = ퟕퟔ. ퟒퟑ°, 풄 = ퟖ. ퟖ or 풎∠푩 = ퟏퟏퟕ. ퟗퟑ°, 풎∠푪 = ퟐퟎ. ퟓퟕ°, 풄 = ퟑ. ퟏퟖ

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2. A surveyor is working at a river that flows north to south. From her starting point, she sees a location across the river that is ퟐퟎ° north of east from her current position, she labels the position 푺. She moves ퟏퟏퟎ⁡feet north and measures the angle to 푺 from her new position, seeing that it is ퟑퟐ° south of east. a. Draw a picture representing this situation.

b. Find the distance from her starting position to 푺. 퐬퐢퐧(ퟓퟐ°) 퐬퐢퐧(ퟓퟖ°) = ퟏퟏퟎ 풙 풙 ≈ ퟏퟏퟖ. ퟒ

The distance is about ퟏퟏퟖ⁡퐟퐭.

c. Explain how you can use the procedure the surveyor used in this problem (called triangulation) to calculate the distance to another object.

Calculate the angle from the starting point to the object. Travel a distance from the starting point, and again calculate the angle to the object. Create a triangle connecting the starting point, the object, and the new location. Use that distance traveled and the angles to find the distance to the object.

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3. Consider the triangle pictured below.

푩푫 풄 퐬퐢퐧(풎∠푩푨푫) Use the law of sines to prove the generalized angle bisector theorem, that is, = . (Although this 푫푪 풃 퐬퐢퐧(풎∠푪푨푫) is called the generalized angle bisector theorem, we do not assume that the bisector of ∠푩푨푪 intersects side 푩푫 풄 푩푪̅̅̅̅ at 푫. In the case that 푨푫̅̅̅̅ is an angle bisector, then the formula simplifies to = .) 푫푪 풃 풄 풃 a. Use the triangles 푨푩푫 and 푨푪푫 to express and as a ratio of sines. 푩푫 푫푪 풄 퐬퐢퐧(풎∠푩푫푨) = 푩푫 퐬퐢퐧(풎∠푩푨푫) 풃 퐬퐢퐧(풎∠푪푫푨) = 푫푪 퐬퐢퐧(풎∠푪푨푫)

b. Note that angles 푩푫푨 and 푨푫푪 form a linear pair. What does this tell you about the value of the sines of these angles?

Since the angles are supplementary, the sines of these values are equal.

c. Solve each equation in part (a) to be equal to the sine of either ∠푩푫푨 or ∠푨푫푪. 풄 퐬퐢퐧(풎∠푩푨푫) ⋅ = 퐬퐢퐧(풎∠푩푫푨) 푩푫 풃 퐬퐢퐧(풎∠푪푨푫) ⋅ = 퐬퐢퐧(풎∠푪푫푨) 푫푪

d. What do your answers to parts (b) and (c) tell you?

The answers tell me that the two equations written in part (c) are equal to each other.

e. Prove the generalized angle bisector theorem.

From part (d), we have 풄 풃 퐬퐢퐧(풎∠푩푨푫) ⋅ = 퐬퐢퐧(풎∠푪푨푫) ⋅ 푩푫 푫푪 Dividing both sides by 퐬퐢퐧(풎∠푪푨푫) ⋅ 풃, and multiplying by 푩푫, we get 푩푫 풄 퐬퐢퐧(풎∠푩푨푫) = . 푫푪 풃 퐬퐢퐧(풎∠푪푨푫)

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4. As an experiment, Carrie wants to independently confirm the distance to Alpha Centauri. She knows that if she measures the angle of Alpha Centauri and waits ퟔ months and measures again, then she will have formed a massive triangle with two angles and the side between them being ퟐ⁡퐀퐔 long. a. Carrie measures the first angle at ퟖퟐ°⁡ퟖ′ퟐퟒ. ퟓ′′ and the second at ퟗퟕ°⁡ퟓퟏ′ퟑퟒ′′. How far away is Alpha Centauri according to Carrie’s measurements?

The third angle would be ퟏ. ퟓ′′. ퟖ ퟐퟒ. ퟓ ퟏ. ퟓ 퐬퐢퐧 (ퟖퟐ + + ) 퐬퐢퐧 ( ) ퟔퟎ ퟑퟔퟎퟎ = ퟑퟔퟎퟎ 풂 ퟐ 풂 ≈ ퟐퟕퟐ ퟒퟑퟔ About ퟐퟕퟐ, ퟒퟑퟔ⁡퐀퐔.

b. Today, astronomers use the same triangulation method on a much larger scale by finding the distance between different spacecraft using radio signals and then measuring the angles to stars. Voyager 1 is about ퟏퟐퟐ⁡퐀퐔 away from Earth. What fraction of the distance from Earth to Alpha Centauri is this? Do you think that measurements found in this manner are very precise? ퟏퟐퟐ Voyager 1 is about , or ퟎ. ퟎퟎퟎퟒ the distance of Earth to Alpha Centauri. Depending on how far away ퟐퟕퟔ ퟑퟔퟒ the object being measured is, the distances are fairly precise on an astronomical scale. One 푨푼 is almost ퟗퟑ million miles, which is not very precise.

5. A triangular room has sides of length ퟑ. ퟖ⁡퐦, ퟓ. ퟏ 퐦, and ퟓ. ퟏ⁡퐦. What is the area of the room? ퟏ.ퟗ ퟑ. ퟖ 퐜퐨퐬(휽) = Since the room is isosceles, the height bisects the side of length at a right angle. We get ퟓ.ퟏ; therefore, 휽 ≈ ퟔퟖ. ퟏퟐퟕ°. ퟏ ⋅ ퟑ. ퟖ ⋅ ퟓ. ퟏ ⋅ 퐬퐢퐧(ퟔퟖ. ퟏퟐퟕ°) ≈ ퟖ. ퟗퟗ ퟐ The area of the room is approximately ퟖ. ퟗퟗ⁡퐦ퟐ.

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6. Sara and Paul are on opposite sides of a building that a telephone pole fell on. The pole is leaning away from Paul at an angle of ퟓퟗ° and towards Sara. Sara measures the angle of elevation to the top of the telephone pole to be ퟐퟐ°, and Paul measures the angle of elevation to be ퟑퟒ°. Knowing that the telephone pole is about ퟑퟓ⁡퐟퐭. tall, answer the following questions. a. Draw a diagram of the situation.

b. How far apart are Sara and Paul?

We can use law of sines to find the shared side between the two triangles and then again with the larger triangle to find the distance: 퐬퐢퐧(ퟐퟐ°) 퐬퐢퐧(ퟓퟗ°) = ퟑퟓ 풓 풓 ≈ ퟖퟎ. ퟎퟖퟔ

Then we use this value in the larger triangle: 퐬퐢퐧(ퟏퟐퟒ°) 퐬퐢퐧(ퟑퟒ°) = 풙 ퟖퟎ. ퟎퟖퟔ 풙 ≈ ퟏퟏퟖ. ퟕퟑퟑ

They are about ퟏퟏퟖ. ퟕ⁡퐟퐭. away from each other.

c. If we assume the building is still standing, how tall is the building? 풚 퐬퐢퐧(ퟓퟗ°) = ퟑퟓ 풚 ≈ ퟑퟎ. ퟎퟎퟏ

The building is about ퟑퟎ⁡퐟퐭. tall.

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