Grade 11 Essential Mathematics Trigonometry

Grade 11 Essential Mathematics

Unit 5: Trigonometry

Page 1

Grade 11 Essential Mathematics Trigonometry

Unit 5: Trigonometry

Introduction: This unit deal with Pythagorean Theorem, and the trigonometric ratios of sine, cosine, and tangent. You will be using these to solve word problems as well. Trigonometry is based on the relationship between the measure of the angles and the lengths of the sides of a right angle triangle. These skills are necessary in occupations such as carpentry, aviation and astronomy. You will also learn how to solve 2D and 3D triangles.

Assessment:

o o o o o o o

Lesson 1 Assignment: Problem Solving using Pythagorean Theorem Lesson 2 Assignment: SOH CAH TOA Lesson 3 Assignment: Word Problems using Trigonometric Ratios Lesson 4 Assignment: Angles of Elevation and Declination Lesson 5 Assignment: Two Triangle Problems Putting It Together – Trigonometry Lesson 6 Assignment: Solving 3-D Triangle Problems

Unit 5: Test Trigonometry

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Grade 11 Essential Mathematics Trigonometry

LESSON 1: PYTHAGOREAN THEOREM The Pythagorean Theorem is for right angle triangle only.

Pythagorean Theorem:

c

a

Hypotenuse

𝑎2 + 𝑏 2 = 𝑐 2 b The hypotenuse is the side that is always opposite (across from) the 90° angle. This is always side C in the Pythagorean Theorem. Example: Determine the length of the missing for the following triangle. Solution:

x

3

𝑥 2 = 32 + 42 𝑥 2 = 9 + 16 𝑥 2 = 25 𝑥 = √25 = 5

4

Example: Determine the length of the missing side for each of the following triangle. Solution: 102 = 𝑥 2 + 82 100 = 𝑥 2 + 64 100 − 64 = 𝑥 2 80 = 𝑥 2 𝑥 = √80 = 8.9 Example: Determine the length of the missing side for each of the following triangle. Solution: 152 = 𝑥 2 + 72 225 = 𝑥 2 + 49 225 − 49 = 𝑥 2 360 = 𝑥 2 𝑥 = √360 = 18.9

10 x

8

15

x 7

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Grade 11 Essential Mathematics Trigonometry

Word Problems using the Pythagorean Theorem Example: Scott wants to swim across a river that is 400m wide. He plans to swim directly across the river but ends up 100m downstream because of the current. How far did he actually swim? Solution: Step 1: Draw a diagram for the right angle triangle:

400 m

x

100 m Step 2: Label the sides of the triangle. Pay attention to where the hypotenuse is!!!

400 m

x HYP

100 m 𝑥 2 = 4002 + 1002 𝑥 2 = 160000 + 10000 𝑥 2 = 170000 𝑥 = √170000 = 412.3𝑚

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Grade 11 Essential Mathematics Trigonometry

Example: To get from point A to point B you must avoid walking through a building. To avoid the building, you walk 14m south and 25m east. How many metres would you have saved had the building not been there? Solution: Must walk 14 + 25 = 39𝑚 x

14 m

142 + 252 = 𝑥 2 196 + 625 = 𝑥 2 821 = 𝑥 2 𝑥 = √821 = 28.65𝑚

25 m

If was able to walk through the building would only have to walk 28.65 m. So would save 39 − 28.65 = 10.35𝑚 Example: The foot of a 6m ladder is placed 2m from the base of a building. How far up the building does that ladder reach? Solution: 6m x

𝑥 2 + 22 = 62 𝑥 2 + 4 = 36 𝑥 2 = 36 − 4 𝑥 2 = 32 𝑥 = √32 = 5.7𝑚

2m

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Grade 11 Essential Mathematics Trigonometry

Example: Calculate x to the nearest tenth of a metre. 12 m 7m

x

4m

Solution: 𝑦 2 + 72 = 122 𝑦 2 + 49 = 144 𝑥 2 = 144 − 49 𝑥 2 = 95 𝑥 = √95 = 9.75𝑚

𝑥 2 + 42 = 9.752 𝑦 2 + 16 = 95 𝑥 2 = 95 − 16 𝑥 2 = 79 𝑥 = √79 = 8.89𝑚

Page 6

Grade 11 Essential Mathematics Trigonometry

Curriculum Outcomes: 11E4.Develop a spatial sense related to triangles.

Lesson 1 Assignment: Problem Solving using Pythagorean Theorem

See your teacher for Lesson 1 Assignment

Page 7

Grade 11 Essential Mathematics Trigonometry

LESSON 2: TRIGONOMETRY The three trig functions are SIN, COS, and TAN SOH CAH TOA Stands for: sin 𝜃 =

𝑜𝑝𝑝 ℎ𝑦𝑝

cos 𝜃 =

𝑎𝑑𝑗 ℎ𝑦𝑝

tan 𝜃 =

𝑜𝑝𝑝 𝑎𝑑𝑗

When we are using the trigonometric functions, we need to label the sides of the given right angle triangle so that we know what number goes where in our formulas. The three sides are: 1. Hypotenuse: is the side of the triangle that is always opposite the 90° angle 2. Opposite: is the side that is opposite (across from) the given angle, we never use the 90° angle to find the opposite side 3. Adjacent: is the side that is adjacent (next to) the given angle we never use the 90° angle to find the adjacent side Example: Label the sides of the following triangle.

θ Solution: The side that is across from the 90° angle is the Hypotenuse. The side that is across from the angle 𝞱 is the Opposite and the side that is next to the given angle 𝞱 is the Adjacent side.

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Grade 11 Essential Mathematics Trigonometry

Using the calculator:

First, we need to make sure that our calculator is set in Degrees. To do this, check the screen on the calculator, it should have a D, DEG, or a DRG. If it has any of the following we need to change it to Degrees: R, RAD, G, GRAD. To change the settings on the calculator look for a button that had DRG on it. Press it until you see D, DRG or DEG on the screen of your calculator. This will make sure that your calculator is in Degrees. It is important that we do this since all of the angles in this unit are measured in degrees, if your calculator is set in something different all of your answers will be incorrect. Depending on your calculator you will either enter the trigonometric function first or the number first. Example: 𝑠𝑖𝑛 45 = 0.7071 Example: 𝑐𝑜𝑠 73 = 0.2924 Example: 𝑡𝑎𝑛 54 = 1.3764

We can also use the functions to determine the size of the angle. These are called the Inverse Trig Functions. These are found on the calculator and not as their own buttons. To use these functions we must press the INV, Shift, or 2nd button on the calculator Example: 𝑐𝑜𝑠  = 0.7 Solution:

𝑠𝑖𝑛−1 𝜃

𝑐𝑜𝑠 −1 𝜃

𝑡𝑎𝑛−1 𝜃

𝜃 = cos −1 0.7 𝜃 = 45°

Example: 𝑠𝑖𝑛  = 0.15 Solution:

𝜃 = sin−1 0.15 𝜃 = 8.6°

Example: 𝑡𝑎𝑛  = 6.2 Solution:

𝜃 = tan−1 6.2 𝜃 = 81° Page 9

Grade 11 Essential Mathematics Trigonometry

Trigonometric Ratios Example: Determine the length of the missing side. 15.2 7

320

x

x 0

45 Solution:

sin 45 =

𝑥

sin 32 =

7 𝑥

0.7071 = 7

𝑥 15.2 𝑥

𝐶𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥

0.5299 = 15.2

(0.7071)(7) = 𝑥

(0.5299)(15.2) = 𝑥

4.95 = 𝑥

8.05 = 𝑥

Example: Solve for the hypotenuse. x x

220

4.6

7 0

45 Solution:

sin 45 = 0.7071 =

4.6 𝑥 4.6 𝑥

(0.7071)(𝑥) = 4.6 4.6

sin 22 =

𝑥 = 0.7071 = 6.5

7 𝑥 7

𝐶𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦

0.3746 = 𝑥

𝐷𝑖𝑣𝑖𝑑𝑒 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥

(0.3746)(𝑥) = 7 7

𝑥 = 0.3746 = 18.7

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Grade 11 Essential Mathematics Trigonometry

Example: Solve for the missing angle. 21 15

θ

8

16 θ

Solution:

sin 𝜃 =

8

sin 𝜃 =

15

16 21

sin 𝜃 = 0.5333

sin 𝜃 = 0.7619

𝜃 = sin−1 0.5333 = 32.2°

𝜃 = sin−1 0.7691 = 50.3°

Example: Determine the length of the missing side. 19 7.5

320 x

450 x Solution:

cos 45 =

𝑥

cos 32 =

7.5 𝑥

0.7071 = 7.5

𝐶𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥

𝑥 19

𝑥

0.848 = 19

(0.7071)(7.5) = 𝑥

(0.848)(19) = 𝑥

5.3 = 𝑥

16.11 = 𝑥

Page 11

Grade 11 Essential Mathematics Trigonometry

Example: Solve for the hypotenuse. x x

220 7.9

450 7.6 Solution:

cos 45 = 0.7071 =

7.6

cos 22 =

𝑥

7.6 𝑥

(0.7071)(𝑥) = 7.6

7.9 𝑥

7.9

𝐶𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦

0.9272 =

𝐷𝑖𝑣𝑖𝑑𝑒 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥

(0.9272)(𝑥) = 7.9

7.6

𝑥

7.9

𝑥 = 0.7071 = 5.4

𝑥 = 0.9272 = 8.5

Example: Solve for the missing angle. θ

21 15

8

θ 16

Solution:

cos 𝜃 =

8 15

cos 𝜃 =

16 21

cos 𝜃 = 0.5333

cos 𝜃 = 0.7619

𝜃 = cos −1 0.5333 = 57.7°

𝜃 = cos −1 0.7691 = 40.4°

Page 12

Grade 11 Essential Mathematics Trigonometry

Example: Determine the length of the missing side. 320

x

x

10.2

0

45 7 Solution:

tan 45 =

𝑥

tan 32 =

7

𝑥

1 = 7.5

𝑥 10.2 𝑥

𝐶𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥

0.6249 = 19

(1)(7.5) = 𝑥

(0.6249)(19) = 𝑥

7.5 = 𝑥

11.8 = 𝑥

Example:. Solve for the missing side. 220

16 7

450 Solution:

tan 45 = 1=

x

16

16 𝑥

16 1

tan 22 =

𝑥

(1)(𝑥) = 16 𝑥=

x

= 16

7 𝑥

7

𝐶𝑟𝑜𝑠𝑠 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦

0.4040 = 𝑥

𝐷𝑖𝑣𝑖𝑑𝑒 𝑡𝑜 𝑠𝑜𝑙𝑣𝑒 𝑓𝑜𝑟 𝑥

(0.4040)(𝑥) = 7 7

𝑥 = 0.4040 = 17.3

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Grade 11 Essential Mathematics Trigonometry

Example: Solve for the missing angle. θ

8

16

12

θ 5

Solution:

tan 𝜃 =

8 15

tan 𝜃 =

16 21

tan 𝜃 = 0.5333

tan 𝜃 = 0.7619

𝜃 = tan−1 0.5333 = 86.2°

𝜃 = tan−1 0.7691 = 37.3°

Page 14

Grade 11 Essential Mathematics Trigonometry

Curriculum Outcomes: 10E2.TG.3. Solve problems that require the manipulation and application of primary trigonometric ratios 10E2.TG.2.Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent)

Lesson 2 Assignment: SOH CAH TOA

See your teacher for Lesson 2 Assignment

Page 15

Grade 11 Essential Mathematics Trigonometry

LESSON 3: USING THE TRIG RATIOS TO SOLVE WORD PROBLEMS Example: Suppose a kite handle has 15ft of kite string. If the wind picks up and the kite string makes an angle with the ground of 56°, what is the height of the kite? Solution: 𝑥

sin 56 = 15 𝑥

15 ft

0.829 = 15

x ft

(0.829)(15) = 𝑥 12.4 𝑓𝑡 = 𝑥

56°

𝑇ℎ𝑒 𝑘𝑖𝑡𝑒 𝑖𝑠 12.4 𝑓𝑒𝑒𝑡 𝑜𝑓𝑓 𝑡ℎ𝑒 𝑔𝑟𝑜𝑢𝑛𝑑

Example: A wire supports a tower and forms an angle of 57° with the ground. The wire is attached to the ground at a point that is 8.5m away from the base of the tower. a. at what height is the wire attached to the tower? b. how long is the wire? Solution:

𝑡𝑎𝑛57 =

𝑥 8.5 𝑥

1.5399 = 8.5

ym

(1.5399)(8.5) = 𝑥

xm

13.1𝑚 = 𝑥

57° 8.5 cos 57 = 𝑦 0.5446 = 𝑦=

𝑡ℎ𝑒 𝑤𝑖𝑟𝑒 𝑖𝑠 𝑎𝑡𝑡𝑎𝑐ℎ𝑒𝑑 13.1𝑚 𝑜𝑓𝑓 𝑡ℎ𝑒 𝑔𝑟𝑜𝑢𝑛𝑑

8.5 m

8.5 𝑦

8.5 = 15.6𝑚 0.5446

𝑡ℎ𝑒 𝑤𝑖𝑟𝑒 𝑖𝑠 15.6𝑚 𝑙𝑜𝑛𝑔

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Grade 11 Essential Mathematics Trigonometry

Example: A 6.1m ladder leans against a wall. The angle formed by the ladder and the wall is 71°. a. how far is the base of the ladder from the wall? b. how far up the wall does the ladder reach? Solution: 𝑥

sin 71 = 6.1

71°

𝑥

0.9455 = 6.1 (0.9455)(6.1) = 𝑥

6.1m

5.8 𝑚 = 𝑥

ym

𝑇ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟 𝑖𝑠 5.8 𝑚 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙

cos 71 =

𝑦 6.1

xm

(0.3256)(6.1) = 𝑦 𝑦 = 1.98 𝑚 𝑡ℎ𝑒 𝑙𝑎𝑑𝑑𝑒𝑟 𝑟𝑒𝑎𝑐ℎ𝑒𝑠 2 𝑚 𝑢𝑝 𝑡ℎ𝑒 𝑤𝑎𝑙𝑙 Example: A truck travels 6km up a mountain road. The change in height is 1.25km. What is the angle of the road? Solution:

sin 𝜃 =

1.25 km

1.25 6

sin 𝜃 = 0.2083

6 km

𝜃 = sin−1 0.2083

𝞱

𝜃 = 12°

Page 17

Grade 11 Essential Mathematics Trigonometry

Curriculum Outcomes: 10E2.TG.3. Solve problems that require the manipulation and application of primary trigonometric ratios 10E2.TG.2.Demonstrate an understanding of primary trigonometric ratios (sine, cosine, tangent)

Lesson 3 Assignment: Word Problems using Trigonometric Ratios

See your teacher for Lesson 3 Assignment

Page 18

Grade 11 Essential Mathematics Trigonometry

LESSON 4: ANGLE OF ELEVATION AND DECLINATION Definition: 1. Angle of Elevation: is the angle formed up from the horizontal 2. Angle of Declination: is the angle formed down from the horizontal

Example: From a point 8m from the base of a building, you measure the angle up to the top of the building from eye level and find that it is 50°. If you are 1.2m tall, how tall is the building? Solution:

tan 50 = x

1.19 =

y

50

𝑥 8

𝑥 8

(1.19)(8) = 𝑥

8m 1.2m

𝑥 = 9.5𝑚

𝑆𝑜 𝑡ℎ𝑒 ℎ𝑖𝑒𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑖𝑠 9.5 + 1.2 = 10.7 𝑚 𝑡𝑎𝑙𝑙

Page 19

Grade 11 Essential Mathematics Trigonometry

Example: The highest point on a cliff is 90m above the shore. From the top of the cliff, a surveyor measures the angle of declination to a boat in the lake to be 42°. How far away from shore is the boat? Solution: x

tan 42 =

42 90m

0.9 = 𝑥=

90 𝑥

90 𝑥

90 0.9

𝑥 = 100𝑚

x

Example: From a point 5m from the base of a tree, you measure the angle of elevation using a 1.5m tall instrument to be 39°. How tall is the tree? Solution:

tan 39 = 0.81 =

y X

𝑦 5

𝑦 5

𝑦 = (5)(0.81) = 4.05𝑚

39 1.5m 5m

𝑆𝑜 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑟𝑒𝑒 𝑖𝑠 𝑋 = 4.05 + 1.5 = 5.55𝑚 𝑡𝑎𝑙𝑙.

Page 20

Grade 11 Essential Mathematics Trigonometry

Example: A 5m tall lighthouse sits at the top of a 30m cliff and the top of the lighthouse is 35m above sea level. The angle of depression to a fishing boat is 24°. The angle of depression to a second boat past the fishing boat is 16°. How far apart are the two boats? Solution:

24

y

16

35m 35m

35m

x 𝑡𝑜 𝑓𝑖𝑛𝑑 ℎ𝑜𝑤 𝑓𝑎𝑟 𝑎𝑝𝑎𝑟𝑡 𝑡ℎ𝑒 𝑏𝑜𝑎𝑡𝑠 𝑎𝑟𝑒 𝑓𝑟𝑜𝑚 𝑒𝑎𝑐ℎ𝑜𝑡ℎ𝑒𝑟 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐𝑙𝑎𝑐𝑢𝑙𝑎𝑡𝑒: 𝑋 = 𝑌 − 𝑍 𝐹𝑖𝑟𝑠𝑡 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑘𝑛𝑜𝑤 ℎ𝑜𝑤 𝑓𝑎𝑟 𝑎𝑤𝑎𝑦 𝑏𝑜𝑎𝑡 1 𝑖𝑠 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑐𝑙𝑖𝑓𝑓 35 tan 24 = 𝑧 0.445 = 𝑧=

35 𝑧

35 = 78.7𝑚 0.445

𝑵𝒆𝒙𝒕, 𝒘𝒆 𝒄𝒂𝒍𝒄𝒖𝒍𝒂𝒕𝒆 𝒕𝒉𝒆 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 𝒃𝒐𝒂𝒕 𝟐 𝒊𝒔 𝒇𝒓𝒐𝒎 𝒕𝒉𝒆 𝒄𝒍𝒊𝒇𝒇

𝒚=

𝐭𝐚𝐧 𝟏𝟔 =

𝟑𝟓 𝒚

𝟎. 𝟐𝟖𝟕 =

𝟑𝟓 𝒚

𝟑𝟓 = 𝟏𝟐𝟏. 𝟗𝒎 𝟎. 𝟐𝟖𝟕

𝑿 = 𝟏𝟐𝟏. 𝟗 − 𝟕𝟖. 𝟕 = 𝟒𝟑. 𝟐𝒎 𝒔𝒐 𝒕𝒉𝒆 𝒃𝒐𝒂𝒕𝒔 𝒂𝒓𝒆 𝟒𝟑. 𝟐𝒎 𝒂𝒑𝒂𝒓𝒕 Page 21

Grade 11 Essential Mathematics Trigonometry

Curriculum Outcomes: 11E4.Develop a spatial sense related to triangles.

Lesson 4 Assignment: Angles of Elevation and Declination

See your teacher for Lesson 4 Assignment

Page 22

Grade 11 Essential Mathematics Trigonometry

LESSON 5: SOLVING TWO-TRIANGLE PROBLEMS Example: Nolan works as a tour guide in Rankin Inlet and is often asked to take pictures for tourists standing beside the statue of an inuksuk. If Nolan’s friend Michael is 187cm tall and is standing beside the statue, how tall is the inuksuk? Nolan has measured two angles, one to the top of Michael’s head is 26° and the other to the top of the inuksuk is 50°.

Solution:

tan 26 =

187

0.4877 = x

187

187 𝑦

(𝑦)(0.4877) = 187 (𝑦)(0.4877)

50°

𝑦

0.4877

187

= 0.4877

187

𝑦 = 0.4877 = 383.4𝑐𝑚

26°

y

tan 50 =

𝑆𝑜 𝑁𝑜𝑙𝑎𝑛 𝑖𝑠 𝑠𝑡𝑎𝑛𝑑𝑖𝑛𝑔 383.4𝑐𝑚 𝑎𝑤𝑎𝑦 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑠𝑡𝑎𝑡𝑢𝑒

𝑥 383.4

1.1918 =

𝑥 383.4

(1.1918)(383.4) = 𝑥 456.9𝑐𝑚 = 𝑥 𝑆𝑜 𝑡ℎ𝑒 𝑠𝑡𝑎𝑡𝑢𝑒 𝑖𝑠 456.9𝑐𝑚 𝑡𝑎𝑙𝑙.

Page 23

Grade 11 Essential Mathematics Trigonometry

Example: A camp instructor leads a group of students on a canoe trip across a lake. They leave Half Moon Bay (H) and paddle 7.8km to Jarvis Bay (J). They then turn 69° and paddle 5.1km to a beach. This beach, B, is 7.6km across the lake from Half Moon Bay. At what angle should they turn in order to where they started?

Solution: Even though the diagram looks as though we can cut line segment BJ in half, we cannot assume that this is actually true. Our diagram might not be accurate so we cannot make this assumption.

sin 69 =

𝑦 7.8

0.9336 =

𝑦 7.8

(0.9336)(7.8) = 𝑦 7.28𝑘𝑚 = 𝑥

sin 𝜃 =

7.28 7.6

sin 𝜃 = 0.9579 𝜃 = sin−1 0.9579 = 73°

𝑇ℎ𝑒𝑦 𝑠ℎ𝑜𝑢𝑙𝑑 𝑡𝑢𝑟𝑛 𝑎𝑡 𝑎𝑛 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 73° Page 24

Grade 11 Essential Mathematics Trigonometry

Example: Two office towers are 50m apart. From the top of the shorter tower to the top of the taller towers a worker measures the angle of elevation to be 25° and the angle of depression to the base of the taller tower to be 35°. Determine the height of each of the towers? Solution: Remember that both angle of elevation and depression are measured off of a Horizontal line. We must include this horizontal line from the spot where the angles are measured. The observer is standing on the roof of the shorter office building and it is from this point that the angles are measured so here is where the horizontal line is drawn. Angle of Elevation of 25° is UP from the horizontal and angle of depression of 35° is Down from the horizontal. We want to know the height of both of the towers. We can calculate the height of the shorter tower right away: 𝑥 50 𝑥 0.7002 = 50 (0.7002)(50) = 𝑥 tan 35 =

y 25° 35°

𝑥 = 35𝑚 50m

z 𝑇ℎ𝑒 𝑠ℎ𝑜𝑟𝑡𝑒𝑟 𝑏𝑢𝑖𝑙𝑑𝑖𝑛𝑔 𝑖𝑠 35𝑚

x

x

50m

𝐭𝐚𝐧 𝟐𝟓 =

𝒚 𝟓𝟎

𝟏. 𝟏𝟗𝟏𝟖 =

𝒚 𝟓𝟎

(𝟏. 𝟏𝟗𝟏𝟖)(𝟓𝟎) = 𝒚 𝟐𝟑. 𝟑𝒎 = 𝒚

𝒛 = 𝒙 + 𝒚 = 𝟑𝟓 + 𝟐𝟑. 𝟑 = 𝟓𝟖. 𝟑𝒎

𝒕𝒉𝒆 𝒕𝒂𝒍𝒍 𝒆𝒓 𝒃𝒖𝒊𝒍𝒅𝒊𝒏𝒈 𝒊𝒔 𝟓𝟖. 𝟑𝒎

Page 25

Grade 11 Essential Mathematics Trigonometry

Curriculum Outcomes: 11E4.TG.1. Solve problems that involve two and three right triangles

Lesson 5 Assignment: Two Triangle Problems

See your teacher for Lesson 5 Assignment

Page 26

Grade 11 Essential Mathematics Trigonometry

See your teacher for Assignment “Putting It Together – Trigonometry”

Page 27

Grade 11 Essential Mathematics Trigonometry

LESSON 6: SOLVING 3-D TRIANGLE PROBLEMS Examples: Determine the length of AY in the diagram below.

Solution: 𝐼𝑛 𝑜𝑟𝑑𝑒𝑟 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑙𝑎𝑐𝑘 𝑑𝑜𝑡𝑡𝑒𝑑 𝑙𝑖𝑛𝑒 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑒 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑋 𝑆𝑖𝑛𝑐𝑒 𝑤𝑒 𝑑𝑜𝑛′ 𝑡 ℎ𝑎𝑣𝑒 𝑎𝑛𝑦 𝑎𝑛𝑔𝑙𝑒𝑠 𝑤𝑒 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑢𝑠𝑒 𝑃𝑦𝑡ℎ𝑎𝑔𝑜𝑟𝑒𝑎𝑛 𝑇ℎ𝑚 𝑎2 + 𝑏 2 = 𝑐 2 122 + 92 = 𝑥 2 144 + 81 = 𝑥 2 225 = 𝑥 2 𝑥 = √225 = 15𝑖𝑛

82 + 15 = 𝑐 2 64 + 225 = 𝑐 2 289 = 𝑐 2 𝑐 = √289 = 17𝑖𝑛

Page 28

Grade 11 Essential Mathematics Trigonometry

Example: Carl and Devon are rock climbing instructors. They need to determine the height of the peak they will be climbing. From where Carl is standing it is 51° up to the top of the cliff they will be climbing. Devon is 105m away from Carl. The angle between base camp and Carl is measured by Devon to be 78°. How high is the cliff that they will be climbing?

Solution:

x 78° 105m 𝑤𝑎𝑛𝑡 𝑡𝑜 𝑘𝑛𝑜𝑤 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑖𝑓𝑓— 𝐴𝐵 𝑇𝑜 𝑓𝑖𝑛𝑑 𝐴𝐵 𝑤𝑒 𝑓𝑖𝑟𝑠𝑡 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑘𝑛𝑜𝑤 𝑡ℎ𝑒 𝑙𝑒𝑛𝑔𝑡ℎ 𝑜𝑓 𝑋 𝑥

tan 78 = 105 𝑥 105 (4.7046)(105) = 𝑥 4.7046 =

493.99 𝑚 = 𝑥 𝑊𝑒 𝑐𝑎𝑛 𝑛𝑜𝑤 𝑢𝑠𝑒 𝑥 = 494 𝑚 𝑡𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒 𝑡ℎ𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑙𝑖𝑓𝑓 𝑦 tan 51 = 494 𝑦 1.2349 = 494 (1.2349)(494) = 𝑦 𝑦 = 610 𝑚 Page 29

Grade 11 Essential Mathematics Trigonometry

Example: Anne works in a lighthouse that is 160 feet above sea level. She sees a boat south east from the lighthouse and a second boat that is south west of the lighthouse. She measures the angle of depression to the first boat measures to be 24° and the angle of depression to the second boat to be 38°. How far apart are the two boats?

Solution:

160ft

a

b

x 𝑙𝑒𝑡𝑠 𝑙𝑜𝑜𝑘 𝑎𝑡 𝑒𝑎𝑐ℎ 𝑏𝑜𝑎𝑡 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑒𝑙𝑦. 𝑅𝑒𝑚𝑒𝑏𝑒𝑟 𝑤𝑒 𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝐷𝑒𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑠𝑜 𝑤𝑒 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑑𝑟𝑎𝑤 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑙𝑖𝑛𝑒 𝑓𝑟𝑜𝑚 𝑤ℎ𝑒𝑟𝑒 𝑡ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑖𝑠 𝑚𝑒𝑎𝑠𝑢𝑟𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑡𝑜𝑝 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑔ℎ𝑡ℎ𝑜𝑢𝑠𝑒.

a 24° 160ft 160ft

𝐭𝐚𝐧 𝟐𝟒 =

𝟏𝟔𝟎 𝒂

𝟎. 𝟒𝟒𝟓𝟐 =

𝟏𝟔𝟎 𝒂

(𝒂)(𝟎. 𝟒𝟒𝟓𝟐) = 𝟏𝟔𝟎 a

𝒂=

𝟏𝟔𝟎 = 𝟑𝟓𝟗. 𝟒𝒇𝒕 𝟎. 𝟒𝟒𝟓𝟐

𝑆𝑜 𝑏𝑜𝑎𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 1 𝑖𝑠 359.4 𝑓𝑒𝑒𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑔ℎ𝑡ℎ𝑜𝑢𝑠𝑒

Page 30

Grade 11 Essential Mathematics Trigonometry

b

38°

160ft

160ft

𝐭𝐚𝐧 𝟑𝟖 =

𝟏𝟔𝟎 𝒃

𝟎. 𝟕𝟖𝟏𝟑 =

𝟏𝟔𝟎 𝒃

(𝒃)(𝟎. 𝟕𝟖𝟏𝟑) = 𝟏𝟔𝟎 b

𝟏𝟔𝟎

𝒃 = 𝟎.𝟕𝟖𝟏𝟑 = 𝟐𝟎𝟒. 𝟖𝒇𝒕 𝑆𝑜 𝑏𝑜𝑎𝑡 𝑛𝑢𝑚𝑏𝑒𝑟 2 𝑖𝑠 204.8 𝑓𝑒𝑒𝑡 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑏𝑎𝑠𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑔ℎ𝑡ℎ𝑜𝑢𝑠𝑒 𝑎2 + 𝑏 2 = 𝑐 2 359.42 + 204.82 = 𝑥 2 x

b

129168.36 + 41943.04 = 𝑥 2 171107.57 = 𝑥 2

a

𝑥 = √171107.57 = 413.65 𝑓𝑡

𝑆𝑜 𝑡ℎ𝑒 𝑏𝑜𝑎𝑡𝑠 𝑎𝑟𝑒 413.65 𝑓𝑒𝑒𝑡 𝑎𝑝𝑎𝑟𝑡

Page 31

Grade 11 Essential Mathematics Trigonometry

Curriculum Outcomes: 11E4.TG.1. Solve problems that involve two and three right triangles.

Lesson 6 Assignment: Solving 3-D Triangle Problems

See your teacher for Lesson 6 Assignment

Page 32