What are 3 names for the following angle? Angle GOD Angle DOG Angle O.
Angle Rules
description
Transcript of Angle Rules
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Angles on a straight line add to 180
180a b c
a
bc
d
e
ab
c
Angles round a point add to equal 360
360a b c d e
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a b
c
d
Vertically Opposite Angles are Equal
a b
c d
a
b
c
Vert opp 's
sum add to 180
Angles in a triangle add to equal 180
180a b c
Base angles of an isosceles triangle are equal
Base 's isosc are
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Exterior angle of a triangle equals the sum
of the opposite two interior angles
180
180
a b c
c d
a b c c
b
c c
d
a d
2intExt of sum opp s
a c
b
d
Complementary Angles add to 90o
The complement of 55o is 35o because these add to 90o
Supplementary Angles add to 180o
The supplement of 55o is 125o because these add to 180o
C before S90 before 180
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35
x 35x
2aa 50a
20a
38
x
*
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x
42 x
x
x42
75
*
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123
67x
*
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Rules of Parallel Lines
Corresponding angles of parallel
lines are equal
Alternate angles of parallel
lines are equal
Co-interior angles of parallel
lines add to 180
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We can say 143 cointerior to 37x
37 corresponding to 37
or s on st ln with
y
x
143 Vertically opposite
or s on st ln with
z x
y
37x
y
z
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x
57
27
27 Alt of lns ares° Ð =P
57 27 30
30 Alt of lns ares° Ð =P
So we can say 30 (Alt of lns are )x s Twice= ° Ð =P
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57
x
65
65 , Corresp s lns area= ° Ð =P
65 57 180
180 65 57
58
sum add to 180
x
x
x
a
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36
x
36
opp of a gram are
x
s
= °
Ð =P
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42
72x
72 ,Alt s lns area= ° Ð =P
42 72 180
180 42 72
66
x
x
x
sum add to180
a
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3 5x
4 10x
We make a statement
3 5 4 10 (Corresp s lns are )
15
x x
x
+ = - Ð =
=
P
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No ofsides
Name NoOf
Degrees in
polygon
Each interior angle for regular
polygons(sides are equal)
Sum of exterior angles
3 Triangle
4 Quadrilateral
5 Pentagon
6 Hexagon
etc
12 Dodecagon
1
2
10
180
180×2=360180×3=540180×4=720
180×10=1800
180÷3=60
360÷4=90
540÷5=108
720÷6=120
1800÷12=150
360
360
360
360
360
360
3
4
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Sum of interior anglesEach interior angle
Number of sides
Sum of the angles in a Polygon (No of sides -2) 180
Sum of interior anglesNumber of sides
Each interior angle
For regular polygons only
For ANY polygon
For regular polygons only
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(5 2) 3 no of sides-2
Degrees in the polygon : Degrees in the polygon:
(6 2) 4 no of sides-2
95
135
x
13595
3 180 540
95 95 135 135 540
460 540
540 460
80
x
x
x
x
x
155 130
80
135130
4 180 720
15 130 80 130 135 720
630 720
720 630
90
x
x
x
x
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A regular polygon has 9 sides what is the interior angle?
9 2 180 1260 sum of all angles
1260140 size of each interior angle
9
The sum of all the angles in a polygon is 2340 .If each interior
angle is 156 , how many sides does the polygon have?
234015 no of sides
156
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Eg Interior angle is 150 . Find the number of sides.
Ext Angle 180 150
=30
360No of sides
Ext Angle
360
30
12
This is a regular Polygon
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Similar Triangles
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Similar TrianglesIf triangles are similar:Corresponding side lengths are in proportion. (One triangle is an enlargement of the other)Corresponding angles in the triangle are the same
25m20mx4 m
It doesn’t matter which way round you make the fraction BUT you must do the same for both sides
little little
big big
4
25 204
25205
x
x
x
It is sensible to start with the x so it is on the top
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If the angles of two triangles are the same, they are similar triangles.
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48 m
12 m
15 m
x
y
48 m
12 m
15 mx
Start with unknown on the top
48
15 1248 15
1260
x
x
x m
60 15
45
y
y m
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20 m
2m
1.5m
4.5m
20 m
2m3m
l
2
20 32 20
31
133
l
l
l m
#11
x
x
x
x
x
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Lesson 6
Circle Language and Angle at Centre
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Equal Radii: Two radii in a circle always form an isosceles triangle
Isos , = radii
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76
x
*
37
Base ‘s isos Δ, = radii
x
Base ‘s isos Δ, = radii
Sum of Δ = 180°
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Angle at the centre is twice the angle at the circumference
a
a
aa
2a
2a
2a
2a
2at centre at circum
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Angle on the circumference of a semicircle is a right angle
in semi-circle
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Lesson 7
Tangent is perpendicular to the radius and Angles on
Same Arc are equal
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Tangent is perpendicular to the radius
Tan radius
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Angles on the same arc are equal
‘s On the same arch equal
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35
62
x
y
18
x
y
*
59x
y
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42
x
y85 38 x
y
z
*
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66x
85
y
*
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63x
y
z
Find unknowns and give reasons
*
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43
57
y
x
z
Find unknowns and give reasons
*
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Cyclic Quadrilaterals
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Cyclic Quadrilaterals
ab
c
d
opp s cyclic quad
180
180
opposite angles of a cyclic
quadrilateral add to equal 180
opposite angles of a cyclic
quadrilateral are supplementary
a b
c d
A quadrilateral which has all four vertices on the circumference of a circle is called a Cyclic quadrilateralRule 1:
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Rule 2:
The exterior angle of a cyclic quadrilateral is equal to the opposite interior angle
,ext cyclic quad
a
b
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67x 95
x
140
x
y
*
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37
x
y 34
x y
*
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43
x
y
*
Find unknowns and give reasons
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110
Find unknowns and give reasons
*
x
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Tangents
Tan radius
When two tangents are drawn from a point to a circle, they are the same length
tangents
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25w
x
yz
140
*
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Similar TrianglesIf triangles are similar:Corresponding side lengths are in proportion. (One triangle is an enlargement of the other)Corresponding angles in the triangle are the same
25m20mx4 m
It doesn’t matter which way round you make the fraction BUT you must do the same for both sides
little little
big big
4
25 204
25205
x
x
x
It is sensible to start with the x so it is on the top
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If the angles of two triangles are the same, they are similar triangles.
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48 m
12 m
15 m
x
y
48 m
12 m
15 mx
Start with unknown on the top
48
15 1248 15
1260
x
x
x m
60 15
45
y
y m
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20 m
2m
1.5m
4.5m
20 m
2m3m
l
2
20 32 20
31
133
l
l
l m
#11
x
x
x
x
x
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Revision
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Geometric reasoning revision
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2006 examQUESTION ONEThe diagram shows part of a fence.AD and BC intersect at E.Angle AEB = 48°.Angle BCD = 73°.Calculate the size of angle CDE.
QUESTION TWOThe diagram shows part of another fence.LM = LN.KL is parallel to NM.LM is parallel to KN.Angle LNK = 54°.
Calculate the size of angle LMN.
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2006 exam
The points A, B, C and D lie on a circle with centre O.Angle OAD = 55°.Angle DOC = 68°.Calculate the size of angle ABC.You must give a geometric reason for each step leading to your answer.
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QUESTION THREEThe diagram shows the design for a gate.
AE = 85 cmBE = 64 cmCD = 90 cm Triangles ABE and ACD are similar. Calculate the height of the gate, AD.
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QUESTION FOURThe diagram shows a design for part of a fence.GHIJK is a regular pentagon and EHGF is a trapezium.AB is parallel to CD.Calculate the size of angle EHG.You must give a geometric reason for each step leading to your answer.
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QUESTION FIVE
The diagram shows another fence design.ACDG is a rectangle.Angle CBA = 110°.CG is parallel to DE.DA is parallel to EF.Calculate the size of angle DEF. You must give a geometric reason for each step leading to your answer.
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• In the above diagram, the points A, B, D and E lie on a circle.• AE = BE = BC.• The lines BE and AD intersect at F.• Angle DCB = x°.• Find the size of angle AEB in terms of x.• You must give a geometric reason for each step leading to your
answer.