chapter 6 additional topics: triangles and vectors 6.1 law of sines 6.2 law of cosines 6.3 areas of...
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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
6.1 Law of Sines
Deriving the Law of SinesSolving ASA and AAS casesSolving the ambiguous SSA case
The Law of Sines
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
Using the Law of Sines (AAS case)
Example: Solve this triangle:º - (63º + 38º) = 79º
.1363sin
79sin12
79sin1263sin
79sin
12
63sin
inc
cc
SSA Variations
6.2 Law of Cosines
Deriving the Law of CosinesSolving the SAS caseSolving the SSS case
Law of Cosines
Strategy for Solving the SAS Case
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
Strategy for the SSS Case
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.
6.3 Area of Triangles
Base and height givenTwo sides and included angle givenThree sides given (Heron’s Formula)Arbitrary triangles
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
Three Sides Given
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
6.4 Vectors
Velocity and standard vectorsVector addition and Scalar multiplicationAlgebraic PropertiesVelocity VectorsForce VectorsStatic Equilibrium
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
Vector Addition
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)
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
Algebraic Properties of Vectors
The Dot Product
The dot product of two vectorsAngle between two vectorsScalar component of one vector onto
anotherWork
The Dot Product
Computing Dot Products
Example: Find the dot product of (4,2) and (1,-3)
Solution: (4,2)·(1,-3)=4·1 + 2·(-3) = -2
Angle Between Two Vectors
3.421713
11cos
1713
11cos
)1,4(),3,2(
1
vu
Scalar Component of u on v
67.217
11
)1,4(),3,2(
u
vu
vcomp
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