Post on 21-Jan-2016
description
transcript
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
Chapter 7Applications of Trig and 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?
ASA
SAA
AAA
YES
YES
NO
Congruence Shortcuts
SSS
SAS
SSA
YES
YES
NO
Congruence Shortcuts
Data required for SolvingOblique 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
C
A
R
Given:
AR = ER
EC = AC
Show /_E = /_A
~
~
~
1. AR = ER
2. EC = AC
3. RC = RC
ΔRCE = ΔRCA /_E = /_A
~
~
~
~ ~
Given
Given SSS CPCTC
Reflexive
E
C
A
R
Given:
/_E = /_A
/_ECR = /_ACR
Show AR = ER
~
~
~
2. / ECR = / ACR
3. RC = RC
ΔRCE = ΔRCA AR = ER
~
~
~
~ ~
Given
Given AAS CPCTC
Reflexive
1. /_E = /_A
E
C
A
R
Given:
/_E = /_A
/_ERC = /_ARC
Show AR = ER
~
~
~
2. / ERC = / ARC
3. RC = RC
ΔRCE = ΔRCA AR = ER
~
~
~
~ ~
Given
Given ASA CPCTC
Reflexive
1. /_E = /_A
Law of Sines
In any triangle ABC, with sides a, b, and c,
= , = , =
This can be written in compact form as
= =
asin A
bsin B
asin A
csin C
bsin B
csin C
asin A
csin C
bsin B
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 AcuteNumber of
Possible Triangles SketchCondition Necessary for Case to Hold
0 a<h
1 a=h
1 a>b
2 b>a>h
Ambiguous Case
Number of Possible Triangles
SketchCondition 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,
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
Oblique Triangle Case 1
• One side and two angles AAS or ASA1. Find the remaining angle using the angle
sum formula (A+B+C)=1802. Find the remaining sides using the Law of
Sines
Oblique Triangle Case 2
• Two sides and a non-included angle SSA1. 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 SAS1. 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 SSS1. 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 andthe 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
To find the product of a real number k and a vector A : kA=A+A+…+A (k times)
Example: 3A
Scalar Product
Magnitude and DirectionAngle 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 Vectorsi = <1, 0> j = <0, 1>
i, j Form for VectorsIf 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 ofDot 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
48
60
60
48
v-v
B
A
130à
|v|2 = 482 + 602 – 2(48)(60)cos(130à)|v|2 ≈ 9606.5|v| ≈ 98 newtons
The required angle can be found by subtracting angle CAB from 180à.
C
98 60
sin 130à sin CAB=
CAB ≈ 28à so £ = 180à - 28à =152à
Inclined Application
20à50
x
sin 20à = |AC|/50|AC| ≈ 17 pounds of force
AC