C2 st lecture 8 pythagoras and trigonometry handout
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Lecture 8 - Pythagoras and Trigonometry C2 Foundation Mathematics (Standard Track) Dr Linda Stringer Dr Simon Craik [email protected] [email protected] INTO City/UEA London
Transcript of C2 st lecture 8 pythagoras and trigonometry handout
Lecture 8 - Pythagoras and TrigonometryC2 Foundation Mathematics (Standard Track)
Dr Linda Stringer Dr Simon [email protected] [email protected]
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
