chapter 6 additional topics: triangles and vectors 6.1 law of sines 6.2 law of cosines 6.3 areas of...
TRANSCRIPT
![Page 1: Chapter 6 Additional Topics: Triangles and Vectors 6.1 Law of Sines 6.2 Law of Cosines 6.3 Areas of Triangles 6.4 Vectors 6.5 The Dot Product](https://reader035.vdocuments.mx/reader035/viewer/2022081419/56649e985503460f94b9b565/html5/thumbnails/1.jpg)
Chapter 6Additional Topics: Triangles and
Vectors
6.1 Law of Sines
6.2 Law of Cosines
6.3 Areas of Triangles
6.4 Vectors
6.5 The Dot Product
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6.1 Law of Sines
Deriving the Law of SinesSolving ASA and AAS casesSolving the ambiguous SSA case
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The Law of Sines
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Using the Law of Sines (ASA case)
Example: Solve this triangle:Solution:
º
= 180º - (45.1º + 75.8º) = 59.1º
.42.81.59sin
1.45sin2.10
1.45sin2.101.59sin2.10
1.59sin1.45sin
ina
aa
min. 18 hr. 1about or hr, 3.1mi/hr 130
mi 164
mi.164
12cos)258)(5.97(2)258()5.97(
mi. 97.5 hr. 3/4mi./hr. 130222
d
d
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Using the Law of Sines (AAS case)
Example: Solve this triangle:º - (63º + 38º) = 79º
.1363sin
79sin12
79sin1263sin
79sin
12
63sin
inc
cc
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SSA Variations
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6.2 Law of Cosines
Deriving the Law of CosinesSolving the SAS caseSolving the SSS case
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Law of Cosines
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Strategy for Solving the SAS Case
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Using the Law of Cosines (SAS case)
'40100)45'2034(180
4598.7
'2034sin10sin
98.7
'2034sin
10
sin
mb 98.7'2034cos)10)(9.13(2)10()9.13(
. and , b,for triangle thisSolve :Example
22
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Strategy for the SSS Case
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Navigation
min. 18 hr. 1about or hr, 3.1mi/hr 130
mi 164
mi.16412cos)258)(5.97(2)258()5.97(
mi. 97.5 hr. 3/4mi./hr. 130222
dd
Example: Find how far a plane has flown off course at 12º after flying for ¾ of an hour.
Also, find how much longer the flight will take.
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6.3 Area of Triangles
Base and height givenTwo sides and included angle givenThree sides given (Heron’s Formula)Arbitrary triangles
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Base and Height Given
Example: Find the area of this triangle.
Solution:A = (ab/2) sin q =
½ (8m)(5m) sin 35º
≈ 11.5 m2
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Three Sides Given
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Using Heron’s Formula
Example: Find the area of the triangle with sides a = 12 cm, b = 8 cm, and c = 6 cm.
Solution:
s = (12 + 8 + 6)/2 = 13 cm.
A = √(13(13-12)(13-8)(13-6) =
√(13(1)(5)(7) ≈ 21 cm2
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6.4 Vectors
Velocity and standard vectorsVector addition and Scalar multiplicationAlgebraic PropertiesVelocity VectorsForce VectorsStatic Equilibrium
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Finding a Standard Vector for a Given Geometric Vector
The coordinates (x, y) of P are given by
x = xb – xa = 4 – 8 = -4
y = yb – ya = 5 – (-3) = 8
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Vector Addition
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Scalar Multiplication
Let u = (-5, 3) and v = (4, -6)u + v = (-5 + 4, 3 + (-6)) = (-1, -3)-3 u = -3(-5, 3) = (-3(-5), -3(3)) = (15, -9)
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Unit Vectors
10
1,
10
3)1,3(
10
1
||
1
1013||)1,3(
v.ofdirection in r unit vecto Find :Example
22
vv
u
vv
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Algebraic Properties of Vectors
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The Dot Product
The dot product of two vectorsAngle between two vectorsScalar component of one vector onto
anotherWork
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The Dot Product
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Computing Dot Products
Example: Find the dot product of (4,2) and (1,-3)
Solution: (4,2)·(1,-3)=4·1 + 2·(-3) = -2
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Angle Between Two Vectors
3.421713
11cos
1713
11cos
)1,4(),3,2(
1
vu
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Scalar Component of u on v
67.217
11
)1,4(),3,2(
u
vu
vcomp
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Work
Example: How much work is done by a force F = (6,4) that moves an object from the origin to the point p = (8, 2)?
Solution: w = (6,4)·(8,2) = 56 ft-lb