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Chapter 7
Applications of Trig and Vectors
Section 7.1 Oblique Triangles & Law of Sines
Section 7.2 Ambiguous Case & Law of Sines
Section 7.3 The Law of Cosines
Section 7.4 Vectors and the Dot Product
Section 7.5 Applications of Vectors
Section 7.1 Oblique Triangles &
Law of Sines
• Congruency and Oblique Triangles
• Law of Sines
• Solving using AAS or ASA Triangles
Congruency and Oblique Triangles
• If we use A for angles and S for sides what
are all of the three letter combinations you
could create?
• Which of these can we use to prove the
triangles are congruent?
Congruence Shortcuts
ASA
YES
SAA
YES
NO
AAA
Congruence Shortcuts
SSS
YES
SAS
YES
NO
SSA
Data required for Solving
Oblique Triangles
1. One side and two angles (ASA or AAS)
2. Two sides and one angle not included
between the two sides (ASS). Yep this
one can create more than one triangle
3. Two sides and the angle between them
(SAS)
4. Three sides (SSS)
5. Three angles (AAA) Yep this one only
creates similar triangles.
E
Given:
AR ~
= ER
EC ~
= AC
R
C
Show /_E ~= /_A
A
1.
~
AR = ER
Given
2.
~
EC = AC
Given
3.
Reflexive
~
ΔRCE = ΔRCA
SSS
~
RC = RC
~
/_E = /_A
CPCTC
E
Given:
/_E ~= /_A
~
/_ECR = /_ACR
Show AR ~
= ER
R
C
A
~ /_A
1. /_E =
Given
~
/ ECR = / ACR
2.
Given
3.
~
ΔRCE = ΔRCA
AAS
~
RC = RC
Reflexive
AR ~
= ER
CPCTC
E
Given:
/_E ~= /_A
~
/_ERC = /_ARC
Show AR ~
= ER
R
C
A
~ /_A
1. /_E =
Given
~
/ ERC = / ARC
2.
Given
3.
~
ΔRCE = ΔRCA
ASA
~
RC = RC
Reflexive
AR ~
= ER
CPCTC
Law of Sines
In any triangle ABC, with sides a, b, and c,
a
b
a
c
b
c
=
,
=
,
=
sin A sin B sin A
sin C sin B
sin C
This can be written in compact form as
a
b
c
=
=
sin A
sin B
sin C
Area of a Triangle
In any triangle ABC, the area A is given by
any of the following formulas:
A = ½bc sin A
A = ½ab sin C
A = ½ac sin B
Section 7.2 Ambiguous Case &
Law of Sines
• Description of the Ambiguous Case
• Solving SSA Triangles (Case 2)
• Analyzing Data for Possible Number
Ambiguous Case Acute
Number of
Possible Triangles
Sketch
Condition Necessary for Case to Hold
0
a<h
1
a=h
1
a>b
2
b>a>h
Ambiguous Case
Number of
Possible
Triangles
Sketch
Condition Necessary
for Case to Hold
0
a<b
1
a>b
SSA Cases
• Remember since SSA results in two
possible triangles we must check the
angles supplement as well. So if we find
the angle is 73 then we also have to
check 180 – 73 = 107.
Section 7.3 The Law of Cosines
•
•
•
•
Derivation of the Law of Cosines
Solving SAS Triangles Case 3
Solving SSS Triangles Case 4
Heron’s Formula for the Area of a Triangle
Triangle Side Length Restriction
• In any triangle, the sum of the lengths of
any two sides must be greater than the
length of the remaining side.
Law of Cosines
Law of Cosines
In any triangle ABC, with sides a, b, and c,
= + – 2bc cos A
2
2
2
b = a + c – 2ac cos B
2
2
2
c = a + b – 2ab cos C
2
a
2
b
2
c
Oblique Triangle Case 1
• One side and two angles AAS or ASA
1. Find the remaining angle using the angle
sum formula (A+B+C)=180
2. Find the remaining sides using the Law of
Sines
Oblique Triangle Case 2
• Two sides and a non-included angle SSA
1. Find an angle using the Law of Sines
2. Find the remaining angle using the Angle
Sum Formula
3. Find the remaining side using the Law of
Sines
There may be no triangle or two triangles
Oblique Triangle Case 3
• Two sides and an included angle SAS
1. Find the third side using the Law of Cosines
2. Find the smaller of the two remaining angles
using the Law of Sines
3. Find the remaining angle using the angle
sum formula
Oblique Triangle Case 4
• Three sides SSS
1. Find the largest angle using the Law of
Cosines
2. Find either remaining angle using the Law of
Sines
3. Find the remaining angle using the angle
sum formula
Heron’s Area Formula
If a triangle has sides of lengths a, b, and c,
and if the semi-perimeter is
s= ½(a+b+c)
then the area of the triangle is
A =  s(s-a)(s-b)(s-c)
Section 7.4 Vectors and
the Dot Product
•
•
•
•
•
Basic Vector Terminology
Finding Components and Magnitudes
Algebraic Interpretation of Vectors
Operations with Vectors
Dot Product and the Angle between
Vectors
Basic Terminology
• scalars – quantities involving only
magnitudes
• vector quantities – quantities having both
magnitude and direction
• vector – a directed line segment
• magnitude – length of a vector
• initial point – vector starting point
• terminal point – second point through which
the vector passes
Sum of vectors
To find the sum of two vectors A and B:
A+B
resultant vector
or
Difference of vectors
To find the difference of 2 vectors A and B:
A+(-B)
resultant vector
or
Scalar Product
To find the product of a real number k and a
vector A : kA=A+A+…+A (k times)
Example: 3A
Magnitude and Direction
Angle of a Vector <a,b>
The magnitude of vector u=<a,b> is
given by
|u| = a2 + b2
The direction angle  satisfies
tan  =b/a,
where a ≠ 0.
Horizontal and Vertical
Components
The horizontal and vertical components,
respectively, of a vector u having
magnitude |u| and direction angle  are
given by
 = |u| cos 
and
 =|u| sin 
Vector Operations
For any real numbers a, b, c, d, and k,
<a, b> + <c, d> = <a+c, b+d>
k ·<a, b> = <ka, kb>
If a = <a1, a2>, then -a = <-a1, -a2>
<a, b> - <c, d> = <a, b> + -<c, d>
Unit Vectors
i = <1, 0>
j = <0, 1>
i, j Form for Vectors
If v = <a, b>, then
v = ai + bj
Dot Product
The dot product of the two vectors
u = <a, b> and v = <c, d> is denoted by
u·v, read “u dot v,” and given by
u·v = ac + bd
Properties of the Dot Product
For all vectors u, v, w and real numbers k
u·v=v·u
u ·(v+w) = u·v + u·w
(u+v) ·w= u·w + v·w (ku)·v=k(u·v)=u·(kv)
0·u=0
u · u = |u|2
Geometric Interpretation of
Dot Product
If  is the angle between the two nonzero
vectors u and v, where 0<  <180, then
u·v = |u||v| cos 
Orthogonal Vectors
Two nonzero u and v vectors are orthogonal
vectors if and only if u · v = 0
Section 7.5 Applications of Vectors
• The Equilibrant
• Incline Applications
• Navigation Applications
Equilibrant
A vector that counterbalances the resultant
is called the equilibrant. If u is a vector
then –u is the equilibrant.
u + -u = 0
Equilibrant Force
• Use the law of Cosines
B
130à
v
48
-v
60
A
60
|v|2 = 482 + 602 – 2(48)(60)cos(130à)
|v|2 ≈ 9606.5
|v| ≈ 98 newtons
C
48
The required angle can be
found by subtracting angle CAB
from 180à.
98
60
=
sin 130à sin CAB
CAB ≈ 28à so £ = 180à - 28à =152à
Inclined Application
20à
50
C
A x
sin 20à = |AC|/50
|AC| ≈ 17 pounds of force