m911 les27
TRANSCRIPT
892019 M911 LES27
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ANALYTICAL GEOMETRY 9830802983081
Learning Outcomes and Assessment Standards
Learning Outcome 3 Space shape and measurement
Assessment Standard AS 3(c) and AS 3(a)
The gradient and inclination of a straight line
The equation of a straight line
bull
bull
Overview
In this lesson you will
Discover what is meant by inclination
Use trigonometry to find the inclination of a straight line
Find the angle between two straight lines
Use analytical methods to find the three angles of a triangle
Lesson
The inclination of a straight line
Definition The angle formed by a straight line and the positive direction of the
horizontal
4 OPTIONS
To find this angle we need the concept of a gradient and link it directly to
trigonometry
m =change in y
__
change in x
and tan α = y
_ x
there4 For any line segment AB
tan α = mAB
there4 tan α = y
B ndash y
A
_
x
B
ndash x
A
So α = tanndash1 (mAB
)
y
x
y
x
27LESSON
Ov er v iew
a
B ( x B y
B)
A ( x A yA) θ
A (1 4)
B (ndash3 ndash2)
Lesson
892019 M911 LES27
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Lesson 1 | Algebr a
P age 1P ag e 115
Examples
1 We introduce any horizontal line through AB
Then tan q = mAB
= 4 + 2
_
1 + 3
= 6
_ 4
=3
_ 2
there4 q = tanndash1 ( 3
_ 2
) = 563deg
2 tan q = mPQ
= 6 + 4
_
ndash2 ndash 1
= ndash 10
_ 3
there4 q = tanndash1 ( ndash 10
_ 3
) = ndash733deg
Notice that when tan q lt 0 we are working with the negative angle q So to
get to q we need to add a period of tan which is 180deg
So q = qprime + 180deg = ndash733deg + 180deg = 1067deg
3 Find the size of the angles between the following lines
a y = ndash x + 4 and y = 1
_ 2
x + 3
b y = ndash3 x ndash 4 and y = ndash x
c y = 8 x and y = 2 x + 3
a For L1 tan a
2 lt 0
So a2 = tanndash1(ndash1) + 180deg
= ndash45deg + 180deg
= 135deg
a2prime = 45deg
For L2 a
1 tanndash1 ( 1
_ 2
) + 266deg
The acute angle between the lines will be 45deg + 266deg = 716deg
The obtuse angle between them is 180deg ndash 716deg = 1084deg
b For L1 m = ndash1
q1 = tanndash1(ndash1) + 180deg
= ndash45deg + 180deg
there4 q1 = 135deg
and q1prime = 45deg
For L2 m = ndash3
q2 = tanndash1(ndash3) + 180deg
= ndash716deg + 180deg
q2 = 1084deg
Thus obtuse angle 1084deg + 45 = 1534deg
θ
θrsquo
Q(ndash2 6)
P(1 ndash4)
θ
θrsquo
Q(ndash2 6)
P(1 ndash4)
To use this concept
effectively always
draw a diagram
L1
L2
y = ndash3 x ndash 4
y = ndash x
q1
q1
q2
a1
a2
a2
L1L
2
y = x + 3
y = ndash x + 4
1
2
892019 M911 LES27
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Acute angle 266deg
c L1 m = 8 ndash q
1 = tanndash1(8) = 828deg
L2 m = 2 ndash q
2 = tanndash1(2) = 634deg
there4 ang between lines (acute) = 828 ndash 634deg
= 194deg
The obtuse angle will be = 180deg ndash 194deg
= 1606deg
4 A(ndash1 5) B(2 3) and C(ndash6 ndash1) are co-ordinates of the vertices of ABC
Find the size of the angles
Draw a picture
Tip Find α β and θ the use geometry
mAC
= 6
_ 5
mBC
= 4
_ 8
mAB
= 2
_
ndash3
tan α = 6
_ 5
tan β = 1
_ 2
tan θ = ndash 2
_ 3
α = 502deg β = 266deg θ = 1463deg
By geometry ^ C = α ndash β
^ C = 236deg
^ A = θ ndash α
^ A = 961deg
^ B1 = 603 (ltrsquos in )
Letrsquos find the inclination of each side
First
AC m =5 + 1
_
ndash1 + 6
=6
_
5
there4 q = tanndash1 ( 6
_ 5
) = 502deg
AB m = 5 ndash 3
_
ndash1 ndash 2 = ndash 2
_ 3
there4 b = tanndash1 ( ndash 2
_ 3
) + 180deg
= 1463deg
BC m = 3 + 1
_
2 + 6 = 1
_ 2
a = tanndash1 ( ndash 1
_ 2
) = 266deg
^ C = q ndash a = 502deg 266deg
^ C = 236deg
A(ndash1 5)
B(2 3
C(ndash6 ndash1)
βα
θ
q1
q2
L2
L1
y = 2 x + 3
y = 8 x
1463deg
A(ndash1 5)
B(2 3)
502deg
266deg
266deg
E xample
892019 M911 LES27
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892019 M911 LES27
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6 ^ A = 45deg Find the gradient of ℓ2
7 Find the gradient of AB
8 Calculate the gradient of OC
9 P(ndash1 4) Q(2 2) and
R(ndash6 ndash1) are co-ordinates of PQR find the angles of the triangle
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 25
Lesson 1 | Algebr a
P age 1P ag e 115
Examples
1 We introduce any horizontal line through AB
Then tan q = mAB
= 4 + 2
_
1 + 3
= 6
_ 4
=3
_ 2
there4 q = tanndash1 ( 3
_ 2
) = 563deg
2 tan q = mPQ
= 6 + 4
_
ndash2 ndash 1
= ndash 10
_ 3
there4 q = tanndash1 ( ndash 10
_ 3
) = ndash733deg
Notice that when tan q lt 0 we are working with the negative angle q So to
get to q we need to add a period of tan which is 180deg
So q = qprime + 180deg = ndash733deg + 180deg = 1067deg
3 Find the size of the angles between the following lines
a y = ndash x + 4 and y = 1
_ 2
x + 3
b y = ndash3 x ndash 4 and y = ndash x
c y = 8 x and y = 2 x + 3
a For L1 tan a
2 lt 0
So a2 = tanndash1(ndash1) + 180deg
= ndash45deg + 180deg
= 135deg
a2prime = 45deg
For L2 a
1 tanndash1 ( 1
_ 2
) + 266deg
The acute angle between the lines will be 45deg + 266deg = 716deg
The obtuse angle between them is 180deg ndash 716deg = 1084deg
b For L1 m = ndash1
q1 = tanndash1(ndash1) + 180deg
= ndash45deg + 180deg
there4 q1 = 135deg
and q1prime = 45deg
For L2 m = ndash3
q2 = tanndash1(ndash3) + 180deg
= ndash716deg + 180deg
q2 = 1084deg
Thus obtuse angle 1084deg + 45 = 1534deg
θ
θrsquo
Q(ndash2 6)
P(1 ndash4)
θ
θrsquo
Q(ndash2 6)
P(1 ndash4)
To use this concept
effectively always
draw a diagram
L1
L2
y = ndash3 x ndash 4
y = ndash x
q1
q1
q2
a1
a2
a2
L1L
2
y = x + 3
y = ndash x + 4
1
2
892019 M911 LES27
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Acute angle 266deg
c L1 m = 8 ndash q
1 = tanndash1(8) = 828deg
L2 m = 2 ndash q
2 = tanndash1(2) = 634deg
there4 ang between lines (acute) = 828 ndash 634deg
= 194deg
The obtuse angle will be = 180deg ndash 194deg
= 1606deg
4 A(ndash1 5) B(2 3) and C(ndash6 ndash1) are co-ordinates of the vertices of ABC
Find the size of the angles
Draw a picture
Tip Find α β and θ the use geometry
mAC
= 6
_ 5
mBC
= 4
_ 8
mAB
= 2
_
ndash3
tan α = 6
_ 5
tan β = 1
_ 2
tan θ = ndash 2
_ 3
α = 502deg β = 266deg θ = 1463deg
By geometry ^ C = α ndash β
^ C = 236deg
^ A = θ ndash α
^ A = 961deg
^ B1 = 603 (ltrsquos in )
Letrsquos find the inclination of each side
First
AC m =5 + 1
_
ndash1 + 6
=6
_
5
there4 q = tanndash1 ( 6
_ 5
) = 502deg
AB m = 5 ndash 3
_
ndash1 ndash 2 = ndash 2
_ 3
there4 b = tanndash1 ( ndash 2
_ 3
) + 180deg
= 1463deg
BC m = 3 + 1
_
2 + 6 = 1
_ 2
a = tanndash1 ( ndash 1
_ 2
) = 266deg
^ C = q ndash a = 502deg 266deg
^ C = 236deg
A(ndash1 5)
B(2 3
C(ndash6 ndash1)
βα
θ
q1
q2
L2
L1
y = 2 x + 3
y = 8 x
1463deg
A(ndash1 5)
B(2 3)
502deg
266deg
266deg
E xample
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 45
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 55
ge 118
6 ^ A = 45deg Find the gradient of ℓ2
7 Find the gradient of AB
8 Calculate the gradient of OC
9 P(ndash1 4) Q(2 2) and
R(ndash6 ndash1) are co-ordinates of PQR find the angles of the triangle
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 35
ge 116
Acute angle 266deg
c L1 m = 8 ndash q
1 = tanndash1(8) = 828deg
L2 m = 2 ndash q
2 = tanndash1(2) = 634deg
there4 ang between lines (acute) = 828 ndash 634deg
= 194deg
The obtuse angle will be = 180deg ndash 194deg
= 1606deg
4 A(ndash1 5) B(2 3) and C(ndash6 ndash1) are co-ordinates of the vertices of ABC
Find the size of the angles
Draw a picture
Tip Find α β and θ the use geometry
mAC
= 6
_ 5
mBC
= 4
_ 8
mAB
= 2
_
ndash3
tan α = 6
_ 5
tan β = 1
_ 2
tan θ = ndash 2
_ 3
α = 502deg β = 266deg θ = 1463deg
By geometry ^ C = α ndash β
^ C = 236deg
^ A = θ ndash α
^ A = 961deg
^ B1 = 603 (ltrsquos in )
Letrsquos find the inclination of each side
First
AC m =5 + 1
_
ndash1 + 6
=6
_
5
there4 q = tanndash1 ( 6
_ 5
) = 502deg
AB m = 5 ndash 3
_
ndash1 ndash 2 = ndash 2
_ 3
there4 b = tanndash1 ( ndash 2
_ 3
) + 180deg
= 1463deg
BC m = 3 + 1
_
2 + 6 = 1
_ 2
a = tanndash1 ( ndash 1
_ 2
) = 266deg
^ C = q ndash a = 502deg 266deg
^ C = 236deg
A(ndash1 5)
B(2 3
C(ndash6 ndash1)
βα
θ
q1
q2
L2
L1
y = 2 x + 3
y = 8 x
1463deg
A(ndash1 5)
B(2 3)
502deg
266deg
266deg
E xample
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 45
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 55
ge 118
6 ^ A = 45deg Find the gradient of ℓ2
7 Find the gradient of AB
8 Calculate the gradient of OC
9 P(ndash1 4) Q(2 2) and
R(ndash6 ndash1) are co-ordinates of PQR find the angles of the triangle
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 45
892019 M911 LES27
httpslidepdfcomreaderfullm911-les27 55
ge 118
6 ^ A = 45deg Find the gradient of ℓ2
7 Find the gradient of AB
8 Calculate the gradient of OC
9 P(ndash1 4) Q(2 2) and
R(ndash6 ndash1) are co-ordinates of PQR find the angles of the triangle