Lecture 8 - Pythagoras and TrigonometryC2 Foundation Mathematics (Standard Track)

Dr Linda Stringer Dr Simon Craikl.stringer@uea.ac.uk s.craik@uea.ac.uk

INTO City/UEA London

Triangles

I A triangle has 3 sides and3 internal angles.

I The sum of the internalangles is 180◦ (degrees).

I α+ β + γ = 180α β

γ

Types of triangle

I If all three sides are the same lengthit is called an equilateral triangle. Inan equilateral triangle all of theangles are 60◦.

I If only two sides are of the samelength it is called an isoscelestriangle. In an isoceles triangle, twoof the angles are equal.

I If no two sides are of the samelength it is called a scalene triangle.

4 cm

4 cm 4 cm

60◦ 60◦

60◦

6 cm 6 cm

80◦ 80◦

Right-angled triangles

hypotenuse

I An angle of 90◦ is a right-angle.I If one of the angles in a triangle is 90◦ then we call it a

right-angled triangle.I In a right-angled triangle, the side opposite the right-angle

is called the hypotenuse. It is always the longest side.

Pythagoras’ theorem

Pythagoras’ theorem states that in a right-angled triangle

a2 + b2 = hyp2

a

bhyp

The square of the hypotenuse is equal to the sum of thesquares of the other two sides.

Pythagoras

I Pythagoras’ theorem is attributed to Pythagoras, a Greekmathematician/philosopher in 6th century B.C.

I Evidence suggests it was known to the Chinese andBabylonians well before this

Example

Consider the following right-angled triangle

3

4c

What is the length c (the hypotenuse)?

c2 = 32 + 42

= 9 + 16 = 25c =

√25 = 5

Example - your turn

Consider the following right-angled triangle

1

2c

What is the length c? Give your answer in surd form

Example - your turn

Consider the following right-angled triangle

3

7c

What is the length c? Give your answer to 1 decimal place

Example

Consider the following right-angled triangle

12

b13

What is the length b?

132 = 122 + b2

169 = 144 + b2

25 = b2

b =√

25 = 5

Example - your turn

Consider the following right-angled triangle

5

b8

What is the length b? Give your answer in surd form

Proof of Pythagoras’ theorem

I Take a right angled triangle:

a

bc

B

A

I As it is a right-angled triangle A + B = 90.

Proof of Pythagoras’ theorem

I Make 4 copies of thetriangle and arrange into asquare.

a

bc

B

A

a

b

cB

A

a

b c

B

A

a

b

cB

A

I The central area is asquare as A + B = 90.

I The centre square hasarea c2.

Proof of Pythagoras’ theorem

I Rotate the triangle to form2 rectangles.

a

bc

B

A

a

b

cB

A

A

BB

A

a

a

bb

I The area outside thetriangles remains thesame.

I However, it is nowmade up of 2 squares.One is of area a2 andthe other is of area b2.

a2 + b2 = c2

The trigonometric ratios: Sine, Cosine and Tangent

I Take a right angled triangle:

adjacent

oppositehypotenuse

A

I We define three functions:

sin(A) =opphyp

, cos(A) =adjhyp

, tan(A) =oppadj

Your turn

Use your calculator to find the following ratios. Fortrigonometric functions, always use surd form or 3 d.p.sin(10)sin(20)sin(30)sin(40)sin(90)

Example

Consider the following triangle

b8

35

What is the length b?

sin(35) = opphyp = b

8b = 8× sin(35)

= 4.6 to 1 d.p.

Example

Consider the following triangle

b4 30

What is the length b?

cos(30) = adjhyp = b

4b = 4× cos(30)

= 2√

3

Your turn

Use your calculator to find the following angles. Give youranswers to 1 d.p.sin−1(0.760)sin−1(1

2)cos−1(0)cos−1(

√3

2 )tan−1(1)tan−1(.5)

Example

Consider the following triangle

27

A

What is sin(A) ?sin(A) = opp

hyp = 27

What is angle A ? Give your answer to 1 d.p.

A = sin−1(27)

= 16.6◦

Example

Consider the following triangle

3

6 A

What is the angle A?

sin(A) = opphyp = 3

6

= 12

A = sin−1(12)

= 30◦

A trigonometric identityIf A is any angle, then

cos2(A) + sin2(A) = 1

ProofI Consider an angle A.I Build a right-angled triangle with angle A.

a

bc A

I We have cos(A) = bc and sin(A) = a

cI It follows that

cos2(A) + sin2(A) =(

bc

)2+(a

c

)2

= b2+a2

c2

= 1 (by Pythagoras’ Theorem)

Sine rule

I In any triangle we have the following:

asin(A)

=b

sin(B)=

csin(C)

a

bc

B C

A

Example

I Find the length of side A

asin(A) = b

sin(B)a

sin(60) = 7sin(40)

a = 7sin(40) × sin(60)

= 9.4 to 1 d.p. a

7

40

60

Your turn

I Find the length of side A

asin(A) = b

sin(B)

a

12

36

72

Proof of the sine rule

I Draw a line down from the corner A down to the edge BCso it is at right angles.

sin(B) =APc

sin(C) =APb

a

bc

BC

A

PThen c sin(B) = AP = b sin(C) so

bsin(B)

=c

sin(C)

Example

Two polar bears are at the North pole. They are 100m awayfrom each other and directly between them is the north pole.One polar bear has an angle of 20◦ to the top of the pole andthe other 30◦. How tall is the pole?

100m

20 30

Example

We have the triangle

100

bc

20 30

A

P

Example

100

bc

20 30

A

P

I The angle at A is 180− 20− 30 = 130.I We have

csin(30)

=100

sin(130), so c =

100× sin(30)sin(130)

= 65.3

I Then

sin(20) =AP65.3

, so AP = 65.3× sin(20) = 22.3

Cosine rule

I In any triangle we have the following:

a2 = b2 + c2 − 2bc cos(A)

a

bc

B C

A

Example

I Find the angle A

4

56 A42 = 52 + 62 − 2× 5× 6× cos(A)16 = 25 + 36− 60 cos(A)−45 = −60 cos(A)34 = cos(A)A = cos−1(3

4) = 41.4

Proof of the cosine rule

I Draw a line down from the corner A down to the edge BCso it is at right angles.

We have

cos(B) =BPc

cos(C) =PCb a

bc

BC

A

PThen

a = BP + PC = c cos(B) + b cos(C)

soa2 = ac cos(B) + ab cos(C)

Proof of the cosine rule

I By symmetry

a2 = ac cos(B) + ab cos(C)b2 = ba cos(C) + bc cos(A)c2 = ca cos(B) + cb cos(A)

I Then

b2 + c2 = ba cos(C) + bc cos(A) + ca cos(B) + cb cos(A)= a2 + 2bc cos(A)

I We rearrange to get:

a2 = b2 + c2 − 2bc cos(A)

Area of a right-angled triangle

In a right-angled triangle

a

b

Area=12

ab

Example

I What is the area of the triangle?

6

b10

102 = 62 + b2

b2 = 64b = 8Area = 1

2 × 8× 6= 24

Proof of the area of a right-angled triangle

Take two copies of the right-angled triangle

They form a rectangle with area ab. Hence, the area of onetriangle is

12

ab

Area of a triangle

I In any triangle we have the following:

Area =12

ab sin(C)

a

bc

B C

A

Example

I What is the area of the triangle?

5

46 A

52 = 42 + 62 − 2× 4× 6× cos(A)25 = 52− 48 cos(A)−27 = −48 cos(A)9

16 = cos(A)A = cos−1( 9

16) = 55.8Area = 1

2 × 4× 6× sin(55.8)= 9.9 to 1 d.p.

Proof of the area of a triangle

Draw a line down from the corner A down to the edge BC so itis at right angles.

I The area of triangle ACP is

12× CP × PA

I The area of triangle ABP is

12× PB × PA a

bc

BC

A

PThen the total area is (1

2 × CP × PA) + (12 × PB × PA)

As CP + PB = a, we have Area = 12 × a× PA

Then since sin(C) = PAb , it follows that

Area =12

ab sin(C)

FormulaePythagoras’ Theorem

a2 + b2 = hyp2

Trigonometric ratios

sin(A) =opphyp

, cos(A) =adjhyp

, tan(A) =oppadj

Sine rulea

sin(A)=

bsin(B)

=c

sin(C)

Cosine rulea2 = b2 + c2 − 2bc cos(A)

AreaArea =

12

ab sin(C)

Pythagoras’ theorem and the trigonometric ratios apply toright-angled triangles only

PractiseQuestion sheet 8

Course bookAttwood et al (2008) Edexcel AS and A Level ModularMathematics, Core Mathematics 1. Heinemann, PearsonsOther recommended booksCroft and Davison. Foundation Mathematics. 5th edition.Pearson Prentice Hall.Lakin, S. How To Improve your Maths Skills. Pearson.

http://www.mathcentre.ac.uk/resources/uploaded/mc-ty-trigratios-2009-1.pdfmathcentre.ac.ukkhanacademy.orgbbc bitesize standard gradehttp://www.cliffsnotes.com/math/trigonometry

Make up your own triangles