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PYTHAGORAS & TRIGONOMETRYPowerPoint Presentation

PYTHAGORAS & TRIGONOMETRY

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## PowerPoint Slideshow about ' PYTHAGORAS & TRIGONOMETRY' - imogene-murray

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Can only occur in a right angled triangle

hypotenuse

Pythagoras Theorem states:

h

a

h2 = a2 + b2

b

right angle

e.g.

x

9.4 cm

7.65 m

y

11.3 m

8.6 cm

x2 = 7.652 + 11.32

y2 + 8.62 = 9.42

9.42 = y2 + 8.62

square root undoes squaring

smaller sides should always be smaller than the hypotenuse

x2 = 186.2125

- 8.62

y2 = 9.42 – 8.62

- 8.62

x = √186.2125

y2 = 14.4

x = 13.65 m (2 d.p.)

y = √14.4

y = 3.79 cm (2 d.p.)

- Special sets of whole numbers that fit into the Pythagoras equation

e.g. 3, 4 and 5

x 2

- Each of these sets can be multiplied by numbers to find further triples.

6, 8 and 10

5, 12 and 13

x 3

15, 36 and 39

7, 24 and 25

x 5

35, 120 and 125

PYTHAGOREAN APPLICATIONS

e.g. A ladder 5 m long is placed

against the wall. The base of

the ladder is 2 m from the

wall.

Draw a diagram to show this

information and calculate

how high up the wall the

ladder reaches.

Wall (x)

Ladder (5 m)

x2 + 22 = 52

52 = x2 + 22

x2 = 52 - 22

-22

-22

x2 = 21

x = √21

x = 4.58 m (2 d.p.)

Base (2 m)

- Label the triangle as follows, according to the angle being used.

to remember the trig ratios use

SOH CAH TOA

Hypotenuse (H)

Opposite (O)

and the triangles

A

Always make sure your calculator is set to degrees!!

Adjacent (A)

O

A

O

S

H

C

H

T

A

means divide

1. Calculating Sides

means multiply

e.g.

O

7.65 m

x

H

h

T

A

O

O

29°

50°

h = tan50 x 6.5

6.5 cm

A

O

x = sin29 x 7.65

h = 7.75 cm (2 d.p.)

x = 3.71 m (2 d.p.)

S

H

d = 455 ÷ sin32

O

d

H

455 m

d = 858.62 m (2 d.p.)

O

S

H

32°

2. Calculating Angles

-Same method as when calculating sides, except we use inverse trig ratios.

e.g.

B

4.07 m

23.4 mm

H

sin-1 undoes sin

2.15 m

16.1 mm

H

A

O

A

cosB = 2.15 ÷ 4.07

sinA = 16.1 ÷ 23.4

A

O

B = cos-1(2.15 ÷ 4.07)

A = sin-1(16.1 ÷ 23.4)

C

H

S

H

B = 58.1° (1 d.p.)

A = 43.5° (1 d.p.)

Don’t forget brackets, and fractions can also be used

e.g. A ladder 4.7 m long is

leaning against a wall. The

angle between the wall and

ladder is 27°.

Draw a diagram and find the

height the ladder extends up

the wall.

A

Wall (x)

C

H

27°

A

Ladder (4.7 m)

x = cos27 x 4.7

H

x = 4.19 m (2 d.p.)

A

e.g. A vertical mast is held by a

48 m long wire. The wire is

attached to a point 32 m up

the mast.

Draw a diagram and find the

angle the wire makes with

the mast.

C

H

H

A

48 m

cosA = 32 ÷ 48

32 m

A = cos-1(32 ÷ 48)

A

A = 48.2° (1 d.p.)

- Vectors describe a movement (translation).

To describe vectors as a column vector:

- top number describes sideways movement

(negative = left and positive = right)

- bottom number describe up/down movement

(negative = down and positive = up)

e.g. Draw the vector q =

b

vectors can start anywhere

q

e.g. Draw the vector b =

e.g. Write vector CD as a column

vector

e.g. Write vector AB as a column

vector

- Bearings are used to indicate directions

- Are measured clockwise from North

- Must be expressed using 3 digits

(i.e. 000° to 360°)

- Compass directions such as NW give directions but are not bearings

000°

e.g. The compass points and their bearings:

315°

045°

e.g. Draw a bearing of 051°:

NW

NE

N

W

E

270°

090°

51°

SW

SE

225°

135°

e.g. What is the bearing of R from N?

S

N

180°

Bearing

= 180 + 37

= 217°

37°

MAGNITUDE AND BEARINGS OF VECTORS

- The magnitude of a vector is its length and is calculated using Pythagoras

- The direction of a vector (bearing) is calculated using Trigonometry

e.g. Calculate the magnitude and bearing of the vector :

N

Magnitude:

x2 = 42 + 62

x = √42 + 62

x = √52

x = 7.2 units (1 d.p.)

x

O

6

Bearing:

tanA = 6 ÷ 4

A = tan-1(6 ÷ 4)

O

A

A = 56°

Bearing

= 56 + 270

T

A

A

4

= 326°

e.g. A plane flying at 500km/hr heads North. A wind blows from the west

at 50 km/hr.

Find the actual direction the plane ends up heading on.

Also calculate its final speed.

50 km/hr

O

N

Final Speed:

x2 = 502 + 5002

Actual flight path (x)

x = √502 + 5002

500 km/hr

x = √252500

A

A

x = 502.5 km/hr (1 d.p.)

Bearing:

tanA = 50 ÷ 500

A = tan-1(50 ÷ 500)

O

A = 5.7° (1 d.p.)

Bearing

= 006°

T

A

GRID REFERENCE from the west

- Grid referencing is a six figure system of map co-ordinates used to give locations

- The first three figures refer to the horizontal scale.

- The last three figures refer to the vertical scale.

e.g. A location has a grid reference of 295868. How is it found?

On the horizontal scale we look for the 29 and then move 5 tenths further right

On the vertical scale we look for the 86 and then move 8 tenths further up

RAPID NUMBERING

- RAPID stands for Rural Address Property IDentification

- It accurately gives the location of rural properties

- It is based on the distance a property is from the beginning of the road

- The distance is measured in metres with the final measurement being divided by 10. If the property is on the right side, the number is rounded to the nearest even whole number and if it is on the left side, to the nearest odd number.

e.g. What is the RAPID number for a property located 527 metres up the left side of a road?

527 ÷ 10 = 52.7

Left side means we round to nearest odd number

= 53

3D FIGURES from the west

- Pythagoras and Trigonometry can be used in 3D shapes

e.g. Calculate the length of sides x and w and the angles CHE and GCH

A

B

x2 = 52 + 62

x = √52 + 62

G

F

x = √61

x = 7.8 m (1 d.p.)

w

O

Make sure you use whole answer for x in calculation

w2 = 72 + 7.82

D

C

7 m

w = √72 + 7.82

A

x

5 m

w = √110

O

w = 10.5 m (1 d.p.)

H

E

6 m

A

tanCHE = 5 ÷ 6

tanGCH = 7 ÷ 7.8

O

O

CHE = tan-1(5 ÷ 6)

GCH = tan-1(7 ÷ 7.8)

T

A

T

A

CHE = 39.8° (1 d.p.)

GCH = 41.9° (1 d.p.)

1. The Angle Between Two Planes from the west

- is the smallest possible angle between the planes.

- is defined by the rays on each plane perpendicular to the line of intersection.

e.g. Find the angle between the planes CHEB and ABCD

B

C

1. First define the two planes

2. Define the line of intersection

3. Define the rays perpendicular to the line of intersection

F

G

A

4. The angle is located between the two rays

7 m

A

D

5 m

O

tanHCD = 5 ÷ 7

O

E

H

6 m

HCD = tan-1(5 ÷ 7)

T

A

HCD = 35.5° (1 d.p.)

2. Angle Between a Line and a Plane from the west

- is the smallest angle between the line and the projection of that line onto the plane.

e.g. Find the angle between the line BH and plane ABFE

B

C

1. Define the line and plane

F

2. Look towards plane and line

G

x

3. Project line onto the plane

A

4. The angle is located between the line and its projection

7 m

A

D

5 m

E

H

6 m

O

Make sure you use whole answer for x in calculation

First need to find length of projection (x)

x2 = 52 + 72

tanEBH = 6 ÷ 8.6

O

x = √52 + 72

EBH = tan-1(6 ÷ 8.6)

x = √74

T

A

EBH = 34.9° (1 d.p.)

x = 8.6 m (1 d.p.)

NON-RIGHT ANGLED TRIANGLES from the west

1. Naming Non-right Angled Triangles

- Capital letters are used to represent angles

- Lower case letters are used to represent sides

e.g. Label the following triangle

c

A

The side opposite the angle is given the same letter as the angle but in lower case.

B

b

C

a

2. Sine Rule from the west

a = b = c .

SinA SinB SinC

a) Calculating Sides

Only 2 parts of the rule are needed to calculate the answer

e.g. Calculate the length of side p

p = 6 .

Sin52 Sin46

52°

A

× Sin52

p = 6 × Sin52

Sin46

× Sin52

B

46°

6 m

b

p = 6.57 m (2 d.p.)

p

a

To calculate you must have the angle opposite the unknown side.

Re-label the triangle to help substitute info into the formula

b) Calculating Angles from the west

For the statement: 1 = 3 is the reciprocal true?

2 6

Yes as 2 = 6

1 3

Therefore to calculate angles, the Sine Rule is reciprocated so the unknown angle is on top and therefore easier to calculate.

You must calculate Sin51 before dividing by 6 (cannot use fractions)

SinA = SinB = SinC

a b c

a = b = c .

SinA SinB SinC

e.g. Calculate angle θ

Sinθ = Sin51

7 6

θ

A

Sinθ = Sin51 × 7

6

× 7

× 7

B

51°

6 m

b

θ = sin-1( Sin51 × 7)

6

7 m

a

θ = 65.0° (1 d.p.)

To calculate you must have the side opposite the unknown angle

Re-label the triangle to help substitute info into the formula

3. Cosine Rule from the west

-Used to calculate the third side when two sides and the angle between them (included angle) are known.

a2 = b2 + c2 – 2bcCosA

a) Calculating Sides

e.g. Calculate the length of side x

x2 = 132 + 112 – 2×13×11×Cos37

13 m

b

x2 = 61.59

A

37°

x = √61.59

x

x = 7.85 m (2 d.p.)

a

11 m

c

Remember to take square root of whole, not rounded answer

Re-label the triangle to help substitute info into the formula

b) Calculating Angles from the west

- Need to rearrange the formula for calculating sides

CosA = b2 + c2 – a2

2bc

Watch you follow the BEDMAS laws!

e.g. Calculate the size of the largest angle

P

CosR = 132 + 172 – 242

2×13×17

24 m

a

Q

CosR = -0.267

13 m

b

A

R = cos-1(-0.267)

17 m

c

R = 105.5° (1 d.p.)

R

Re-label the triangle to help substitute info into the formula

Remember to use whole number when taking inverse

4. Area of a triangle from the west

- can be found using trig when two sides and the angle between the sides (included angle) are known

Area = ½abSinC

e.g. Calculate the following area

Area = ½×8×9×Sin39

9 m

b

C

Area = 22.7 m2 (1 d.p.)

39°

52°

8 m

a

89°

Re-label the triangle to help substitute info into the formula

Calculate size of missing angle using geometry (angles in triangle add to 180°)

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