unit 2: trigonometry - mr. mcdonald'smrmcdonaldshomepage.weebly.com/uploads/1/3/9/2/... · 1...

1

Math 1201

Unit 2: Trigonometry

Read Building On, Big Ideas, and New Vocabulary, p. 68 text.

Ch. 2 Notes

§2.1 The Tangent Ratio (2.5 classes)

Read Lesson Focus p. 70 text.

Outcomes

1. Define a ratio. See notes

2. Identify the hypotenuse of a right triangle and the opposite and adjacent sides for a given

acute angle in the triangle. P. 71

3. Define the tangent ratio in terms of the sides of a right triangle. P.71

4. Define the angle of inclination of a line or line segment. P. 70

5. Use the tangent ratio to determine the measure of a missing angle in a right triangle. P. 72

Naming the Sides of a Right Triangle for a Given Acute Angle

Recall that the side across from the right angle in a right triangle is called the hypotenuse. If we select

an acute angle of the triangle, we can also name the legs of the triangle with respect to that angle. For

example, with respect to acute angle A, the opposite side is BC and the adjacent side is CA (see right

triangle below).

However, with respect to acute angle B, the opposite side is CA and the adjacent side is BC (see right

triangle below).

hypotenuse opposite

adjacent C

B

A

B

hypotenuse

opposite

adjacent

C A

2

nDef : A ratio is the quotient (divide) of two quantities.

E.g.: 1

2 is a ratio because it is the quotient of 1 and 2.

E.g.: 7

x is a ratio because it is the quotient of x and 7.

E.g.: length of the opposite side

length of the adjacent side is a ratio because it is the quotient of the length of the opposite side

and the length of the adjacent side.

E.g.: length of the opposite side

length of the hypotenuse is a ratio because it is the quotient of the length of the opposite side

and the length of the hypotenuse.

E.g.: length of the adjacent side

length of the hypotenuse is a ratio because it is the quotient of the length of the adjacent side and

the length of the hypotenuse.

The Tangent Ratio for a Given Acute Angle

Use a protractor to carefully find m A in each of the right triangles below, and then carefully measure

the length of the side opposite A and the side adjacent to A in each triangle. Complete the table

which follows. (See c\sketch\tanratio)

A

A

A

3

m A Length of side opposite A

(cm)

Length of side adjacent to A

(cm) Length of side opposite A

Length of side adjacent to A

The measure of angle A was the same in each right triangle. Even though the lengths of the

opposite and adjacent sides were different, what was special about the ratio of the lengths of the

opposite and adjacent sides?

For angle A , the ratio of the lengths of the opposite and adjacent sides was the ____________.

Since the ratio of the lengths of the opposite and adjacent sides was the same for angle A, we give this

ratio a special name. It is called the tangent ratio.

length of the opposite side*************Tangent Ratio ******************

length of the adjacent side

With respect to angle A, we write opposite

tanadjacent

A or just o

tana

A

What was the measure of angle A in our right triangle? ________________

With respect to angle A, what is the ratio of the length of the opposite side and the length of the adjacent

side? ________________

So we can write

tan

E.g.: Use the lengths of the sides of the right triangle below to determine tan D and tan E .

length of the opposite side 12tan

length of the adjacent side 5D

length of the opposite side 5tan

length of the adjacent side 12E

5

12

E

D

4

E.g.: Draw a right triangle such that tan 3A .

OR

E.g.: Which triangle below would have 8

tan15

D ? Explain why.

The second triangle has 8

tan15

D because the ratio the length of the opposite side to the length of the

adjacent side is 8

15. In the first triangle, this ratio is

15

8.

Do #’s 3, 6 (Omit f), p. 75 text in your homework booklet.

Using the Tangent Ratio to Find the Measure of an Angle in a Right Triangle

E.g.: Find m E in the triangle below.

Make sure your G.C. is in degree mode!!!!!!!!!!!!!!!!!!!!!!

Using the tangent ratio we can write

1

15tan

8

15tan 61.93

8

E

m E

8

15

E

D 8

15

D

E

4

12

E

A 1

3

E

A

8

15

D

E

5

E.g.: Find m D in the triangle below.



1

8tan

15

8tan 28.07

15

D

m D

Check that 61.93 28.07 90 180

E.g.: If tan 1.7321X , find m X

1tan 1.7321 60m X

E.g.: If tan 1X , find m X

1tan 1 45m X


1tan 0.06 3.43m X


1tan 19.1 87m X

Do #’s 4, 5 a, b, c, (label each vertex), 8, 13, pp. 75-76 text in your homework booklet.

Using the Tangent Ratio to Find the Angle of Inclination

nDef : The angle of inclination of a line or a segment is the angle that it makes with the horizontal.

8

15

D

E

Angle of Inclination

Horizontal

6

E.g.: Determine the angle of inclination to the nearest hundredth of a degree.


1

5.8tan

12.7

5.8tan 24.55

12.7

A

m A

The angle of inclination is about 24.55 .

Your Turn: Find the angle of inclination in the diagram to the right. [Ans: 31.75 ]

Do #’s 10, 11, 15, 16, p. 76 text in your homework booklet.

Problem Solving Using the Tangent Ratio

E.g.: A support cable is anchored to the ground 5m from the base of a telephone pole. The cable is 19m

long and is attached to the top of the pole. What angle, to the nearest tenth of a degree, does the cable

make with the ground?

Using the Pythagorean Theorem, we can find the height, h, of the pole.

2 2 2

2

2

2

19 5

361 25

361 25

336

336 336

h

h

h

h

h

5.8

A Horizontal 12.7

D

19m

5m

Pole

E

7

Using the tangent ratio,

1

336tan

5

336tan 74.7

5

E

m E

Do #’s 17, 18, 20, p. 77 text in your homework booklet.

8

§2.2 Using the Tangent Ratio to Calculate Lengths (1 class)


Outcomes

1. Compare and contrast direct measurement and indirect measurement. P. 78

2. Use the tangent ratio to determine the measure of a missing side of a right triangle. P. 79-81

3. Describe and illustrate situations in which the Pythagorean Theorem cannot be used to find the

length of a side of a right triangle. See notes

4. Solve a problem that involves indirect and direct measurement, using the trigonometric ratios, the

Pythagorean Theorem and measurement instruments such as a clinometer or metre stick.

nDef : Direct measurement is used when we can easily measure a quantity directly.

E.g.: Finding Ryan’s height using a metre stick.

nDef : Indirect measurement is used when it is too difficult or dangerous to measure a quantity

directly. It usually involves measuring some other quantities (e.g.: lengths and angle measures) directly

and using a calculation to find the measure of the quantity we are interested in.

E.g.: Finding the height of the Eiffel Tower. We might use the length of the shadow of the tower and the

height and shadow length of a much smaller object at the same time of day and use similar triangles and

proportions (Grade 9) to find the tower’s height.

E.g.: In the triangle below, if the diagram was drawn to scale, we could measure angle C directly using

a protractor or we could also find this angle indirectly

using the tangent ratio.

Determining the Length of the Side Opposite a Given Acute Angle

E.g.: Find the value of t in the triangle to the right.

Using the Pythagorean Theorem,

2 2 2

2

2

2

13 12

169 144

169 144

25

25 25 5

t

t

t

t

t

9

E.g.: Find BC in the triangle to the right.

We cannot use the Pythagorean Theorem here because we

only know the length of one side.


tan 406

6tan 40 5.0

BC

BC

E.g.: Find the length of the side opposite the 30angle in the triangle to the right.

We could use the Pythagorean Theorem or the tangent ratio.

Let o be the length of the side opposite the 30angle.


2

2 2

2

2

2

10 5 3

100 75

100 75

25

25 25 5

o

o

o

o

o


tan305 3

5 3 tan30 5

o

o

E.g.: Find o in the triangle to the right.

We could use the Pythagorean Theorem but not the tangent ratio

because we do not know the measure of an acute angle..


2

2 2

2

2

2

13 3

13 9

13 9

4

4 4 2

o

o

o

o

o

10

E.g.: Find x in the triangle to the right.

We could use the tangent ratio but not the Pythagorean

Theorem.


tan6714

14tan67 33

x

x

Do #’s 3, 6, 7, 8, 9 a, 12, pp. 82-83 text in your homework booklet.

Determining the Length of the Side Adjacent to a Given Acute Angle

E.g.: Find the length of the side adjacent to the 30angle in the triangle to the right.

We could use the Pythagorean Theorem or the tangent ratio.

Let a be the length of the side adjacent the 30angle.


2 2 2

2

2

2

10 5

100 25

100 25

75

75 75 25 3 25 3 5 3

a

a

a

a

a


5tan30

tan30 5

tan30 5

tan30 tan30

8.7

a

a

a

a

11

E.g.: Find the distance from the base of the tower to the support wire in the triangle to the right.

Let a represent the distance from the base of the tower to the support

wire.

We could use the tangent ratio but not the Pythagorean Theorem.


70tan 68

tan 68 70

tan 68 70

tan 68 tan 68

28.3m

a

a

a

a

E.g.: Find the width, w, of the river to the nearest tenth of a foot.


tan 79100

100tan 79 514.5ft

w

w

E.g.: Find the height of the building in the diagram to the

right.


tan5036

36tan50 42.9ft

o

w

The height of the building is 42.9 6 48.9ft

Do #’s 4, 10, pp. 82-83 text in your homework booklet.

12

§2.3 Measuring an Inaccessible Height (1 class)


Outcomes

1. Define the angle of elevation. P. 94

2. Construct a clinometer and use it to measure an inaccessible height. pp. 84-85

3. Solve a problem that involves indirect and direct measurement, using the trigonometric

ratios, the Pythagorean Theorem and measurement instruments such as a clinometer or metre

stick.

nDef :The angle of elevation is the angle from the horizontal up to the line of sight when you look at an

object.

nDef : A clinometer is an instrument used to measure the angle elevation.

What is the angle of inclination in the picture above?

How would you indirectly measure the height of the light pole in the parking lot (or the height of the

gym wall) using the clinometer?

13

§2.4 The Sine & Cosine Ratios (2 classes)


Outcomes

1. Use the sine ratio or cosine ratio to determine the measure of a missing angle of a right

triangle. pp. 89-96

The Sine Ratio for a Given Acute Angle

Use a protractor to carefully find m A in each of the right triangles below, then carefully measure the

length of the side opposite A and the length of the hypotenuse in each triangle. Complete the table


m A Length of side opposite A

(cm)

Length of Hypotenuse (cm) Length of side opposite A

Length of hypotenuse

A

A

A

14

The measure of angle A was the same in each right triangle. Even though the length of the

opposite side and hypotenuse were different in each triangle, what was special about the ratio of the

lengths of the opposite side and the hypotenuse?

For angle A , the ratio of the lengths of the opposite and the hypotenuse was the ____________.

Since the ratio of the lengths of the opposite side and the hypotenuse was the same for angle A, we give

this ratio a special name. It is called the sine ratio.

length of the opposite side*************Sine Ratio ******************

length of the hypotenuse

With respect to angle A, we write opposite

sinhypotenuse

A or just o

sinh

A


With respect to angle A, what is the ratio of the length of the opposite side and the length of the

hypotenuse? ________________

So we can write

sin

E.g.: Use the lengths of the sides of the right triangle below to determine sin D and sin E .

length of the opposite side 12sin

length of the hypotenuse 13D

length of the opposite side 5sin

length of the hypotenuse 13E

E.g.: Draw a right triangle such that 1

sin3

E .

OR

5

13 12

E

D

4

12

E

A

1

3

E

A

15


sin17

D ? Explain why.

The first triangle has 8

sin15


hypotenuse is 15

17. In the second triangle, this ratio is

8

17.

The Cosine Ratio for a Given Acute Angle

Use a protractor to carefully find m A in each of the right triangles below, then carefully measure the

length of the side adjacent to A and the length of the hypotenuse in each triangle. Complete the table


17

8

15

E

D

17

8

15

D

E

A

A

A

16

m A Length of side adjacent to A

(cm)

Length of Hypotenuse (cm) Length of side adjacent to A

Length of hypotenuse

The measure of angle A was the same in each right triangle. Even though the length of the

adjacent side and hypotenuse were different in each triangle, what was special about the ratio of the

lengths of the adjacent side and the hypotenuse?

For angle A , the ratio of the lengths of the adjacent and the hypotenuse was the ____________.

Since the ratio of the lengths of the adjacent side and the hypotenuse was the same for angle A, we give

this ratio a special name. It is called the cosine ratio.

length of the adjacent side*************Cosine Ratio ******************

length of the hypotenuse

With respect to angle A, we write adjacent

coshypotenuse

A or just a

cosh

A


With respect to angle A, what is the ratio of the length of the adjacent side and the length of the

hypotenuse? ________________

So we can write

cos

Summary

17

E.g.: Use the lengths of the sides of the right triangle below to determine cosD and cos E .

length of the adjacent side 5cos

length of the hypotenuse 13D

length of the adjacent side 12cos

length of the hypotenuse 13E

E.g.: Draw a right triangle such that 12

cos13

E .

OR


cos17

D ? Explain why.

The first triangle has 15

cos17


hypotenuse is 15

17. In the first triangle, this ratio is

8

17.

Do #’s 4, 9, p. 95 text in your homework booklet.

17

8

15

E

D

17

8

15

D

E

24

10

26

E

A

12

5

13

E

A

5

13 12

E

D

18

Using the Sine Ratio or the Cosine Ratio to Find the Measure of an Angle in a Right Triangle

E.g.: Find m E in the triangle below.


Using the sine ratio we can write

1

15sin

17

15sin 61.93

17

E

m E

E.g.: Find m D in the triangle below.


Using the cosine ratio we can write

1

15cos

17

15cos 28.07

17

D

m D

Check that 61.93 28.07 90 180

E.g.: If cos 0.7321X , find m X

1cos 0.7321 42.9m X

E.g.: If 1

sin2

X , find m X

1 1sin 30

2m X

E.g.: If 1

cos2

X , find m X

1cos 0.5 60m X

E.g.: If cos 1X , find m X

1cos 1 0m X

17

8

15

D

E

17

8

15

D

E

19

E.g.: If sin 1.25X , find m X

1sin 1.25 errorm X

Why do you get an error?

Do #’s 6, 7, 8, 10, p. 95 text in your homework booklet.

Problem Solving Using the Sine & Cosine Ratios

E.g.: What is the value of in the diagram to the right?

1

28sin

31

28sin

31

64.6

E.g.: If the ladder is 5m long and its base is 3m from the wall, what angle does the ladder make with the

ground?

1

3cos

5

3cos

5

53.1

20

Your Turn: Find the angle that the ladder makes with the wall in the diagram below. ANS 30


21

§2.5 Using the Sine & Cosine Ratios to Calculate Length (2 classes)


Outcomes

1. Use the sine ratio or cosine ratio to determine the measure of a missing side of a right

triangle. pp. 97-100

Using the Sine & Cosine Ratios to Find the Length of a Leg of a Right Triangle

E.g.: Find the height of the kite in the diagram to the right.

sin 52100

100sin 52

78.8units

x

x

x

E.g.: Assuming B is a right angle, how far is the ladder

from the wall?

cos6010

10cos60

5m

CB

CB

CB

22

Using the Sine & Cosine Ratios to Find the Length of the Hypotenuse of a Right Triangle

E.g.: Find the value of x in the diagram to the right.

2200sin 21

sin 21 2200

2200

sin 21

6138.9m

x

x

x

x

E.g.: Find the value of x in the diagram to the right.

50cos70

cos70 50

cos70 50

cos70 cos70

50

cos70

146.2ft

x

x

x

x

x

Do #’s 3 b, d, 4 b, d, 5-7, 9, 11, 12 a i), b i), 14, pp. 101-102 text in your homework booklet.

23

§2.6 Applying the Trigonometric Ratios (2 classes)


Outcomes

1. Solve a triangle using the primary trigonometric ratios. pp. 106-109

Solving a Right Triangle Given the Lengths of Two Sides

E.g.: Solve the triangle to the right. (i.e.: Find the lengths of the missing sides and the measures of

the missing angles.)

Using the Pythagorean

Theorem,

2 2 2

2

2

4 7

16 49

65

65

65 8.1in

AB

AB

AB

AB

AB


1

4tan

7

4tan 29.7

7

B

m B

Therefore

180 90 29.7 60.3m A

E.g.: Solve the triangle to the right.

Using the Pythagorean

Theorem,

2 2 2

2

2

2

10 5

100 25

100 25

75

75

75 25 3 5 3cm

BC

BC

BC

BC

BC

BC

Using the sine ratio,

1

5sin

10

5sin 30

10

B

m B

Therefore

180 90 30 60m A

Do # 3, p. 111 text in your homework booklet.

24

Solving a Right Triangle Given the Length of One Sides and the Measure of One Acute Angle

E.g.: Solve the triangle to the right.

180 90 25 65m A


tan 258

8tan 25 3.7m

b

b

Using the cosine ratio,

8cos 25

cos 25 8

cos 25 8

cos 25 cos 25

8.8m

c

c

c

c

180 90 51 39m Y


sin519

9sin51 7.0ft

x

x


cos519

9cos51 5.7ft

y

y

Do #’s 4, 5 a, b, 6 a, c, 12 b, pp. 111-112 text in your homework booklet.

E.g.: A road rises 410m for every 1700m traveled horizontally. What is the angle of inclination of the

road?


1

410tan

1700

410tan 13.6

1700

C

m C

The angle of inclination is about 13.6 .

25

E.g.: For the road above, how far does a car travel horizontally when it travels 1000m along the road?


?cos13.6

1000

? 1000cos13.6 972m

The car will travel about 972m horizontally.


E.g.: A boat leaves port and sails 50km due west to an island. It then sails another 80km due south to a

second island. How far is the boat from its port when it reaches the second island?


2 2 2

2

2

80 50

6400 2500

8900

8900

8900 94.3km

d

d

d

d

d

The boat is about 94.3km from port.

E.g.: For the boat above, what is the angle between the path it took south and the path it would take to

return directly to port?


1

50tan

80

50tan 32.0

80m

The angle of 32.0 .


?

1000m

Port

80km

50km

26

E.g.: Find the perimeter and area of the isosceles trapezoid to the right.


4sin 45

sin 45 4

sin 45 4

sin 45 sin 45

5.66units

h

h

h

h


4tan 45

tan 45 4

tan 45 4

tan 45 tan 45

4.0m

a

a

a

a

The perimeter of the trapezoid is 5.66 8 5.66 4 8 4 35.32units .

The area of the trapezoid is 21 18 16 4 48units

2 2a b h

Your Turn: Find the perimeter and area of the isosceles trapezoid to the right. [14.62, 5.77]


27

§2.7 Solving Problems Involving More than One Right Triangle (2 classes)


Outcomes

1. Solve problems involving more than one right triangle using the primary trigonometric ratios.

pp. 113-117.

2. Define the angle of depression. p. 115

E.g.: Find the value of b to the nearest hundredth in the diagram to the right.

Strategy: Find AB using the tangent ratio then find the value of b using the sine ratio.


tan 4886

86 tan 48

95.5127m

AB

AB

AB


sin5195.5127

95.5127sin51 74.23m

b

b


E.g.: Find the length of segment CD in the diagram to the right.

Strategy: Find BD using the tangent ratio, find BC using the tangent ratio, and subtract BC from BD.


tan338.5

8.5tan33

5.5200km

BD

BD

BD


tan308.5

8.5tan30

4.9075km

BC

BC

BC

Therefore,

5.5200 4.9075 0.6125kmCD

28

E.g.: Find AC in the diagram to the right.

Strategy: Find AD using the tangent ratio, find DC using the tangent ratio and add.


tan 428.0

8.0 tan 42

7.2032cm

AD

AD

AD


tan188.0

8.0 tan18

2.5994cm

DC

DC

DC

Therefore,

7.2032 2.5994 9.8026cmAC

Do # 3 a, c, p. 118 text in your homework booklet.

E.g.: Find m ABD in the diagram to the right.

Strategy: Find m CBD using the tangent ratio, find m ABC using the tangent ratio and subtract.


1

4tan

15

4tan 14.9314

15

CBD

m CBD


1

12tan

15

12tan 38.6598

15

ABC

m ABC

Therefore,

38.6598 14.9314

23.7284

m ABD

Do # 5 a, c, p. 118 text in your homework booklet.

B

C D

A

18

18

42

8.0cm

18

15cm

18

A 4cm

18

8cm

18

B

C D

29

nDef :The angle of depression is the angle from the horizontal down to the line of sight when you look

at an object.

E.g.: Find the height of the taller building in the diagram to the right.

Strategy: Find the side adjacent to the 38 angle (the angle of depression) using the tangent ratio, find

the side opposite the 52 angle using the tangent ratio, and add 42m.


42tan38

tan38 42

4253.7575m

tan38

a

a

a


tan 5253.7575

53.7575tan 52

68.8065m

o

o

o

Therefore, the height of the

building is

42 68.8064 110.8m

Do #’s 8, 9 p. 119 text in your homework booklet.


Do #’s 1 b, 3, 5 a i), 6, 8, 9, 11, 13-19, 20 a, 22, 23, pp. 124-126 text in your homework booklet.

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