chapter 5 introduction to trigonometry chapter 5 fileexample 2: solve for x. = 245x = 45 2 x = x...

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© The Visual Classroom

Chapter 5

Trigonometry


Chapter 5 Introduction to Trigonometry

Getting Ready for Trigonometry

A ratio compares two quantities with the same units.

A rate compares two quantities with different units.

(units are not needed)

(units are needed)

A proportion is the equality of two ratios. a cb d

=

a cb d

=If then ad = bc.


Example 1: A case of 12 drinks costs $4.20. Determine the cost per drink.

Cost/drink = $4.20 ÷ 12 drinks

= $0.35/drink

5 29x

=Example 2: Solve for x.

2 45x =452

x =

22.5x =

Cola

95 2x

=

95 55 2x

× = ×

22.5x =

or


Angle Relationships

Complementary angles

Supplementary angles

aba + b = 90º

a + b = 180º ab

Vertically opposite angles are equal.a b

c

da = b c = d


Isosceles triangle

a b

a = b

Sum of the angles of a triangle

a b

ca + b + c = 180º

Exterior angle of a triangle

a c

ba + b = c


Parallel lines

Alternate angles are equala

bd

c

a = b c = d

Corresponding angles are equal g

fhe = f g = h

e

Co-interior angles are supplementarya

db

ca + b = 180ºc + d = 180º

2


Congruent FiguresCongruent figures have the same size and shape.

A

B C

D

E F

If ∆ABC ≅ ∆DEF then

AB = DEBC = EFCA = FD

∠ABC = ∠DEF∠BCA = ∠EFD∠CAB = ∠FDE

anglessides

(≅ means congruent)


Pythagorean Theorem

A

B

C

a

b

ca2 + b2 = c2

a2 = c2 – b2

b2 = c2 – a2

Example: Find the missing side.

x

12 cm

6 cm

x2 + 62 = 122

x2 = 122 – 62

x2 = 144 – 36

x2 = 108108x = x ≈ 10.4 cm


5.2 Investigating and Comparing TrianglesPart 1: Are 3 sides enough?

A B

C

Construct ∆ABCAB = 3 cmBC = 6 cmAC = 5 cm

3 cm

6 cm

5 cm

Construct ∆XYZXY = 3 cmYZ = 6 cm

YX

Z

3 cm

6 cm6 cm6 cm6 cm

2 sides ?


Do the triangles have the same shape?

80º

60º

40º80º

60º

40ºA B

C

D E

F

16 cm

10 cm

14 cm 24 cm

15 cm

21 cm

What do you notice about the lengths of the sides?

The sides are in the same ratio. 1624

=1015

=1421


Part 2: Are two sides enough?

Construct ∆ABCAB = 7 cm

AC = 5 cm A B

C

7 cm

5 cm

5 cmC

Assume ∠A = 40º

Construct ∆ABCAB = 7 cmAC = 5 cm∠A = 40º

A B

C

7 cm

5 cm

40º


A B

C

7 cm

5 cm

40º

Construct ∆DEFDE = 7 cmDF = 5 cm∠E = 40º

D E7 cm

F

5 cm40º

D E

F

7 cm

5 cm

40º

3


Part 3: Are two angles enough?

Construct ∆ABC∠B = 50º∠C = 60º

A

B CA

B C

50º 60º

60º50º


Construct ∆ABC∠B = 50º∠C = 60ºBC = 4 cm

4 cmB C

A

50º 60º

Construct ∆PQR∠Q = 50º∠R = 60ºPQ = 4 cm 50º

4 cm

60º

P

Q R

Are they identical? No


Congruent Triangles: identical in every waysame size and same shape

∆ABC ≅ ∆XYZ

A

B C

X

Y Z

α

β χ

α

β χ

∠A = ∠X∠B = ∠Y∠C = ∠Z

AB = XYAC = XZBC = YZ


Conditions for congruenceA

B C

X

Y Z

1) SSS

A

B C

X

Y Zβ

2) SAS

A

B C

X

Y Zβ

3) ASA

β

χ χβ


Similar Triangles same shape but different sizes

∆ABC ~ ∆XYZ

B C

A

Y Z

X

β

α

χ β

α

χ

All angles are equal

Sides are proportionalAB AC BCXY XZ YZ

= =


Conditions for similarity

B C

A

Y Z

X

AB AC BCXY XZ YZ

= =

1) SSS~

B C

A

Y Z

X

β β

B C

A

Y Z

X

β χ β χ

2) SAS~AB BCXY YZ

=

3) AA~

4


5.4 Modeling with Similar Triangles

Similar triangles have sides which are in the same ratio.AC 22DF 4

= 5.5= The scale factor (k) is 5.5.

The scale factor can be used to calculate other sides.

Ex. If EF = 3 cm, BC = 5.5 × 3 = 16.5 cm

30º 90º

30º 90º4 cm

22 cmA

B

C

D F

E

3 cm16.5 cm


With a scale model we will be able to determine lengths of triangles that we are not able to measure.

Example: The height of a building.

measure this distance

calculate

© The Visual Classroom © The Visual Classroom

Width of the room.

win

dow

s

door

measure

measure

Calculate the width of the room

© The Visual Classroom © The Visual Classroom

5.5 Slopes and Angles of Ramps

5


horizontal projection or run

rise

Installing a wheelchair ramp.

When a ramp is added to provide accessible entrance, the slope should be not more than 1:12. Handrails should be provided whenever the slope is greater than 1:20 and the vertical access is greater than 15 cm.

surface of ramp

handrail


1. Draw a scale diagram to find the angle of rise for a ramp with a slope of 1:12.

112

Angle? 5º


2. Handrails should be provided whenever the slope is greater than 1:20 and the rise is more than 15 cm.

a) Draw a scale diagram to find the angle of rise for a ramp with a slope of 1:20.

201

b) What is the maximum run the ramp could have without rising more than 15 cm? 300 cm

Angle? 3º


1:12over 200

1:1050.1 to 200

1:515.1 to 50

1:20 to 15

Minimum run

Maximum angle

Maximum slope

Vertical rise (mm)

Maximum Slope for Short Ramps

27º11º6º

5º

0 to 30

75.5 to 250501 to 2000

2400


5.6 Investigating the Connection betweenSlope and Angle


The steepness of the roof of a house is referred to as the pitch of the roof by home builders.

6


There are good reasons why some homes have roofs which have a greater pitch.


The steepness of wheelchair ramps is of great importance to handicapped persons.


Engineers refer to the slope of a road as the grade.


Slopes and Lines

rise

run

slope = runrise

The slope of a line is the steepness of the line.


They often refer to the slope as a percentage.


8100

A grade of 8% would mean for every run of 100 units, there is a rise of 8 units.

7


Determine the slope of the line.

5 mm

12 mm

512

m =

We use the letter m because in French the word for “to go up” is monter.


Slopes and Lines

yx

∆=

∆rise

run


riseslope = run


Slopes and Lines

yx

∆=

∆

rise

run


riseslope = run

The steeper the line, the greater the slope.


Angle of inclination or elevation(positive slope)

Angle of declination or depression

(negative slope)

θ

θ

θ2θ1

Parallel lines have the same slope.

(θ theta)


Finding the slope of a line given the angle of inclination.

80º70º60º50º40º30º20º10º

slopeAngle

10º20º

0.170.36

0.580.841.21.732.755.67


–80º–70º–60º–50º–40º–30º–20º–10º

slopeAngle

–10º–20º

–0.17–0.36

Finding the slope of a line given the angle of inclination.

8


Equations of Lines

1- y = mx + b

2- y = m(x – a)

3- y – y1 = m(x – x1)

Slope y-intercept form

Slope x-intercept form

Slope point form


Example 1: Determine the equation of the line which intersects the x-axis at (5, 0) and makes an angle of 40ºwith the x-axis.

40º

40º m = 0.84

y = m(x – a)

y = 0.84(x – 5)


Example 2: Determine the equation of the line which passes through (3, 4) and makes an angle of 30º with the x-axis.

30º

30º m = 0.58

y – y1 = m(x – x1)

y – 4 = 0.58(x – 3)


Trigonometry: The study of triangles (sides and angles)

physics

surveying

Trigonometry has been used for centuries in the study of:

astronomygeography

engineering


A

B

Cadjacent

hypotenuse

opposite


A

B

C

adjacent

opposite

hypotenuse

9


A

B

C

opposite

adjacent

hypotenuse

oppsinhyp

A =adjcoshyp

A =opptanadj

A =


A

B

Copposite

adjacenthypotenuse

oppsinhyp

B =adjcoshyp

B =opptanadj

B =


A

B

C

opp

adj

hyp

SOH CAH TOA


A

B

C

8

6

10

8sin10

A =6cos

10A =

8tan6

A =

opp

adj

hypSOHCAHTOA


A

B

C

3

4

5

4sin5

B =3cos5

B =4tan3

B =

adj

opp

hypSOHCAHTOA


A

B

C

5

12

13

12sin13

B =5cos

13B =

12tan5

B =

adj

opp

hypSOHCAHTOA

10


sin 21° =

cos 53° =

tan 72° =

0.3584

0.6018

3.0777

Use a calculator to determine the following ratios.


sin A = 0.4142

cos B = 0.6820

tan C = 1.562

∠A = sin-1(0.4142)

∠B = cos-1(0.6820)

∠C = tan-1(1.562)

= 24°

= 47°

= 57°

Determine the following angles (nearest degree).


Determine the following angles (nearest degree).

sin A =

cos B =

tan C =

∠A = sin-1(0.5833)

∠B = cos-1(0.2666)

∠C = tan-1(1.875)

= 36°

= 75°

= 62°

712

415

158

= 0.5833

= 0.2666

= 1.875© The Visual Classroom

A

B

C

a6 cm

Example 1: Determine side a

30º

a = 6 sin 30°

a = 3 cm

a = 6 (0.5)sin

6A a

=

sin 306a

=

hypopp

SOHCAHTOA


A

B

C

50º

b

9 m

40º

Ex. 2: Name two trig ratios that will allow us to calculate side b.

1) cos 409b

=

2) sin 509b

=


Ex. 3: In ∆PQR, ∠Q = 90°.

a) Find sin R if PR = 8 cm and PQ = 4 cm.R

P Q4 cm

8 cm4sin

8R = 1

2=

b) Find cos P if PR = 8 cm and QR = 4 cm.R

P Q

8 cm4 cm

4 3cos8

P =PQ2 = 82 – 42

PQ2 = 64 – 16PQ2 = 48

48PQ =3cos

2P =

4 3PQ =

11


A

B

C

Example 4: Determine side b

55º

b

8 cm

b = 8 tan 55°

b = 11.4 cm

b = 8 (1.428)

tan8

B b=

tan 558b

=

opp

adj

SOHCAHTOA


P

Q

R

12 cm

17 cm

Example 5: Determine the measure of ∠P.

cos P = 0.70588

∠P = 45.1°

∠P = cos–1(0.70588)

12cos17

P =

adjacent

hypotenuse

SOHCAHTOA


PQ

R

q

12 cm

Example 6: Determine the measure of side PR.

q(tan 35°) = 12

q = 17.1 cm

12tan 35q

=

12tan 35

q =

120.7007

q =

35°

opp adjMethod 1


PQ

R

q

12 cm

Example 6: Determine the measure of side PR.

q = 12(tan 55°)

q = 17.1 cm

tan 5512q

= 35°

oppadjMethod 2

∠Q = 90° – 35°

∠Q = 55°

55°

q = 12(1.428)


A

B

C

1: Determine side c

41º

c

8 cm

c = 10.6 cm8cos 41c

=

PQ

R25 cm

2: Determine ∠R

20 cm

20sin25

R =

sin R = 0.8

∠R = 53.1º

cos 41 8c =

8cos 41

c =


5.8 Solving ProblemsEx. 1: From a distance of 20 m from the base a lighthouse the angle of elevation to the top of a lighthouse is 38º. Determine the height of the lighthouse.

38º

htan 3820h

=

h = 20 tan 38°

h = 20 (0.7813)

h = 15.6

The lighthouse is 15.6 m high.

20 m

12


26°

Ex. 2: A ladder is leaning against a wall and makes an angle of 26º with the wall. If the ladder is 3 m long, how far up the wall does the ladder reach?

3 m

x

cos 263x

=

x = 3 cos 26°

x = 3 (0.8988)

x = 2.7

The ladder reaches 2.7 m up the wall.


Ex. 3: A local building code states that the maximum slope for a set of stairs is 68 cm rise for every 100 cm of run.

What is the maximum angle at which the set of stairs can rise?

100

68

θ

68tan θ100

=

tan θ = 0.68

θ = tan–10.68θ = 34.2º

The maximum angle is 34.2º

chapter 5 introduction to trigonometry chapter 5 fileexample 2: solve for x. = 245x = 45 2 x = x...

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