Trigonometry

Note 1: Pythagoras’ Theorem

The longest side is always opposite the right angle and is called the hypotenuse (H).

H

x

A

O

Note 1: Pythagoras’ Theorem

In a right-angled triangle the square of the hypotenuse is equal to the sum of the squares on the other two sides.

H a

b

a2 + b2 = H2

hypotenuse

e.g. Pythagoras’ Theorem

Find the side marked d a2 + b2 = H2

d

d

4 cm

7 cm

42 + d2 = 72

16 + d2 = 49

d2 = 49 - 16

d2 = 33

d = √33

d = 5.74 cm (3sf)

e.g. Pythagoras’ Theorem

A cone has a base diameter of 10 cm and a slant height of 15 cm. What is its vertical height?

a2 + b2 = H2

10 cm

52 + x2 = 152

25 + x2 = 225

x2 = 200

x = √200

x = 14.1 cm (3 sf)

x

5cm

Visual Proof of Pythagorean Theorem

e.g. Pythagoras’ Theorem

Solve for x.

x (x-2)

8

(x-2)2 + 82 = x2

x2 -4x +4 + 64 = x2

-4x +4 + 64 = 0

-4x + 68 = 0

-4x = -68

x = -68

-4 x = 17

GAMMA Pg 386-387 Ex.27 .01

1. Find the length of a diagonal of a rectangular box of length 12cm, width 5cm and height 4 cm.

2. A ship sails 20 km due North and then 35 due East. How far is it from its starting point?

3. The diagonal of a rectangle exceeds the length by 2 cm. If the width of the rectangle is 10 cm, find the length. 24 cm

40.3 km

13.6 cm

4. An aircraft flies equal distances SE and then SW to finish 120 km due South of its starting point. How long is each part of its journey?

84.9 km

Note 1: Trig Ratios (Sine, Cosine, Tangent)

Recall: The longest side is always opposite the right angle and is called the hypotenuse (H).

H

35°

A

O

The side opposite the marked angle of 35° is called the opposite (O)

The other side is called the adjacent (A)

Note 1: Sine, Cosine, Tangent

H 55°

A

O

Now…..the side opposite the marked angle of 55° is called the opposite (O)

The other side is called the adjacent (A)

How we label our triangle depends on which angle we are concerned with.

Note 1: Sine, cosine, tangent

In similar triangles, it is clear that the ratio will be the same in both triangles

30

12 6

4 2

30° 30°

O H

6 = 2 = 1 12 4 2

Opp.

Hyp.

Note 1: Sine, Cosine, Tangent

H θ A

O

sin θ =

SOH CA

H TOA

Three important functions are:

cos θ = tan θ = H

O

H

A

A

O

For any angle x the values for sin x, cos x and tan x can be found using either a calculator or tables

e.g. Label the sides of these triangle as opposite to θ (O), adjacent (A) or hypotenuse.

θ

θ

θ

θ

A D

C B

H

A

H

O

A

O O

H

A

A

H

O

Try these! Write trigonometric ratios (in fraction form) for each of the following triangles

θ

x

α

β

A D

C B

7

5

13

12

3

5 5

9

5

Starter (Extention) - Pythagoras’ Theorem

Solve for x.

25

4 (x + 2)

(x+3)

[4(x+2)]2 + (x+3)2 = 252

GAMMA Pg 389-392 Ex. 27.02 Ex 27.03

(4x + 8)2 + (x+3)2 = 252

(16x2 +64x + 64) + (x2+6x+9) = 625

17x2 +70x + 73 = 625

17x2 +70x -552 = 0

Using quadratic formula or GDC, X = 4

G-Solv x = -8.176, x = 4

Using Technology

A scientific or graphics calculator can be used to obtain accurate values of trig ratios.

Use a calculator to find the value of each of the following correct to 4 decimal places.

a.) sin 30° b.) cos54° c.) tan89°

= 0.5000 = 0.5878 = 57.2900

Using Technology

To find an angle, when you know the ratio of two sides we use the inverse trig functions.

a.) sin θ = 0.1073 b.) cos θ = 0.5454 c.) tan θ = ¾

θ =sin-1 .1073 θ =cos-1 .5454 θ =tan-1 .75

θ = 6.2° θ = 56.9° θ = 36.9°

Note 2: Find side length of a right angled triangle

Find x in the equation cos 20° = 3

x

If the size of one angle and the length of one side of a right angled triangle are given, the length of any other side can be found using:

SOH CA

H TOA

x = 3cos 20°

x = 2.82

multiply both sides of the equation by 3

evaluate using the calculator

20

x

3

e.g. Calculate the length of the labelled sides

29°

7 cm

58°

x

y

s

t

cos 29 = sin 29 = 7

x

7

y

7cos 29 = x 7sin 29 = y

x = 6.122 cm y = 3.394 cm

cos 58 = 50

s

50

tsin 58 =

50cos 58 = s 50sin 58 = t

s = 26.50 m t = 42.40 m

4 sf

x both sides by 7

50 m

e.g. Calculate the length of the labelled side

25.4°

10 cm

z

31.3°

x

A

7.4 cm

tan 25.4° = 10

x

10 tan 25.4 = x

x = 4.75 cm

Z

4.7sin 31.3° =

z sin 31.3 = 7.4

z = 14.2 cm

3 sf

H

A

O

H O z = 7.4

sin 31.3

e.g. Calculate the length of the labelled sides

52°

30 cm

.5 m

78°

x

y

s

t

cos 52 = sin 52 = 30

x

30

y

30cos 52 = x 30sin 52 = y

x = 18.47 cm y = 23.64 cm

cos 78 = 5.0

s

5.0

tsin 78 =

.5 cos 78 = s .5 sin 78 = t

s = 0.1040 m t = 0.4891 m

4 sf

GAMMA Pg 402 Ex 28.01

e.g. Find the length marked x

x

32° 10 cm

38°

a.) Find BD from triangle BDC

b.) Now find x from from triangle ABD

tan 32° = 10

BD

D

A

C

B

10 tan 32 = BD

sin 38° = BD

x

x = BD sin 38°

x = 10 tan 32° sin 38°

x = 3.85 cm (3 sf)

Note 3: Finding an unknown angle

Find x in the equation sin x° = 5

3

If we know the length of any two sides in a right angled triangle, it is possible to calculate the size of the other angles:

SOH CA

H TOA

sin x = 0.6

x = sin-1 .6

x°

3 5

1.) Choose the correct trig formula to use based on what sides are given 2.) Substitute side lengths into formula 3.) Change fraction to a decimal 4.) Work out angle using one of the inverse trig keys

x = 36.9° (1 dp)

e.g. Finding an unknown angle

Find x°

10

7

SOH CA

H TOA

sin x =

x = sin-1 ( )

x°

10 7

x = 44.43° (2 dp)

sin x = 0.7

x°

22

16

tan x = 16

22

tan x = 1.375

x = tan-1 ( )

x = 53.97° (2 dp) 10

7

16

22

(2 dp)

e.g. Finding an unknown angle

Find x°

10

5

SOH CA

H TOA

cos x =

x = cos-1 ( )

x° 10

x = 60.0° (1 dp) GAMMA Pg 404 Ex 28.02

cos x = 0.5

x°

35

12

tan x = 12

35

tan x = 2.9167

x = tan-1 ( )

x = 71.1° (1 dp)

5

10

5

12

35

Starter a.) Use pythagoras’ theorem to find the unknown side. b.) Solve for

8

6

24

25

θ

θ

θ

10

7

Starter - Finding Angles

A ramp is 10 m long. It has been constructed so that it rises to a point 1.2 m above the ground.

a.) Draw a diagram and place the measurements 10 m and 1.2 m on the correct sides.

b.) Use trig to calculate the angle between the ramp and the ground.

10 m 1.2 m

θ sinθ = 10

2.1

θ = sin-1 0.12

θ = 6.9°

e.g. Calculate the height of this regular pyramid

122 +162 = D2

400 = D2

D = 20 m

x2 +102 = 262

x2 = 262 -102

x2 = 262 -102

x2 = 576

x = 24

Gamma Ex28.03 pg 406-409 odd Ex29.01 pg 412-413 odd

x

Note 4: Bearings

A bearing is an angle measured clockwise from North. It is given using 3 digits.

e.g. The bearing of B from A is 052°

The bearing of A from B is 232°

N

N

A

B

52° 232°

4 km

5 km

θ

tan θ = 5

4

= 38.7° = 0.8

θ = tan-1 0.8

= 039° ( or 038.7°)

Remember that bearings always have 3 digits

(between 000° and 360°)

e.g. A ship sails 22 km from A on a bearing of 042°, and a further 30 km on a bearing of 090° to arrive at B. What is the distance and bearing of B from A?

a.) Draw a clear diagram and label all points

b.) Find DE and AD

F

N

42°

B 30 km

A

D E

sin 42° = cos 42° = 22

DE

22

AD

22 sin 42° = DE 22 cos 42° = AD

DE = 14.72 km AD = 16.35 km

e.g. A ship sails 22 km from A on a bearing of 042°, and a further 30 km on a bearing of 090° to arrive at B. What is the distance and bearing of B from A?

a.) Draw a clear diagram and label all points

c.) Using ΔABF,

F

N

42°

B 30 km

A

D E

AB2 = AF2 + BF2 (Pythagoras’ Theorem)

AF = DE + EB

= 14.72 + 30 = 44.72 km

BF = AD = 16.35 km

AB2 = 44.722 + 16.352

AB2 = 2267.2

AB = 47.6 km (3 sf)

e.g. A ship sails 22 km from A on a bearing of 042°, and a further 30 km on a bearing of 090° to arrive at B. What is the distance and bearing of B from A?

a.) Draw a clear diagram and label all points

d.) The bearing of B from A is given by the angle DAB.

F

N

42°

B 30 km

A

D E

<DAB = <ABF tan ABF = =

ABF = tan-1 2.7352

BF

AF

35.16

72.44 = 2.7352

ABF = 69.9°

B is 47.6 km from A on a bearing of 069.9°

NuLake Read pg 188-189 Pg 189-191

44.72 km

16.35 km

T

β

First find the angle, β

tan β = 58

130

β = tan-1 2.2414

β = 66°

A + β = 90° A = 90° − 66° A = 24°

Starter