copyright © 2007 pearson education, inc. slide 10-2 chapter 10: applications of trigonometry;...
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Copyright © 2007 Pearson Education, Inc. Slide 10-2
Chapter 10: Applications of Trigonometry; Vectors
10.1 The Law of Sines
10.2 The Law of Cosines and Area Formulas
10.3 Vectors and Their Applications
10.4 Trigonometric (Polar) Form of Complex Numbers
10.5 Powers and Roots of Complex Numbers
10.6 Polar Equations and Graphs
10.7 More Parametric Equations
Copyright © 2007 Pearson Education, Inc. Slide 10-3
10.2 The Law of Cosines and Area Formulas
• SAS or SSS forms a unique triangle
• Triangle Side Restriction– In any triangle, the sum of the lengths of any two
sides must be greater than the length of the remaining side.
Copyright © 2007 Pearson Education, Inc. Slide 10-4
10.2 Derivation of the Law of Cosines
Let ABC be any oblique triangle drawn with its vertices labeled as in the figure below.
cx
Bcy
B cosandsin
BcxBcy cosandsin The coordinates of point A become (c cos B, c sin B).
Copyright © 2007 Pearson Education, Inc. Slide 10-5
10.2 Derivation of the Law of Cosines
Point C has coordinates (a, 0) and AC has length b.
22 )0sin()cos( BcaBcb
Baccab
BacBBca
BcaBacBcb
cos2
cos2)cos(sin
sincos2cos
222
2222
222222
This result is one form of the law of cosines. Placing A or C at the origin would have given the same result, but with the variables rearranged.
Copyright © 2007 Pearson Education, Inc. Slide 10-6
10.2 The Law of Cosines
The Law of Cosines
In any triangle ABC, with sides a, b, and c,
.cos2
cos2
cos2
222
222
222
Cabbac
Baccab
Abccba
Copyright © 2007 Pearson Education, Inc. Slide 10-7
10.2 Using the Law of Cosines to Solve a Triangle (SAS)
Example Solve triangle ABC if
A = 42.3°, b = 12.9 meters, and
c = 15.4 meters.
Abccba cos2222 3.42cos)4.15)(9.12(24.159.12 222 a
7.1092 ameters47.10a
Copyright © 2007 Pearson Education, Inc. Slide 10-8
10.2 Using the Law of Cosines to Solve a Triangle (SAS)
B must be the smaller of the two remaining angles since it is opposite the shorter of the two sides b and c. Therefore, it cannot be obtuse.
9.12
sin
47.10
3.42sin B
47.10
3.42sin9.12sin
B
7.81180 BAC
0.56B
Caution If we had chosen to find C rather than B, we would not have known whether C equals 81.7° or its supplement, 98.3°.
Copyright © 2007 Pearson Education, Inc. Slide 10-9
10.2 Using the Law of Cosines to Solve a Triangle (SSS)
Example Solve triangle ABC if a = 9.47 feet, b =15.9 feet, and c = 21.1 feet.
Solution We solve for C, the largest angle, first. If cos C < 0, then C will be obtuse.
Cabbac cos2222
ab
cbaC
2cos
222
)9.15)(47.9(2
)1.21()9.15()47.9( 222
34109402.9.109C
Copyright © 2007 Pearson Education, Inc. Slide 10-10
10.2 Using the Law of Cosines to Solve a Triangle (SSS)
Verify with either the law of sines or the law of cosinesthat B 45.1°. Then,
.0.25
9.1091.45180
180
CBA
Copyright © 2007 Pearson Education, Inc. Slide 10-11
10.2 Summary of Cases with Suggested Procedures
Oblique Triangle Suggested Procedure for Solving
Case 1: SAA 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 Suggested Procedure for Solving
Case 2: SSA This is the ambiguous case; 0, 1, or 2 triangles.
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.
If two triangles exist, repeat steps 2 and 3.
Copyright © 2007 Pearson Education, Inc. Slide 10-12
10.2 Summary of Cases with Suggested Procedures
Oblique Triangle Suggested Procedure for Solving
Case 3: 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 Suggested Procedure for Solving
Case 4: 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.
Copyright © 2007 Pearson Education, Inc. Slide 10-13
10.2 Area Formulas
• The law of cosines can be used to derive a formula for the area of a triangle given the lengths of three sides known as Heron’s Formula.
Heron’s Formula
If a triangle has sides of lengths a, b, and cand if the semiperimeter is
),(21
cbas
.))()(( csbsass A
Then the area of the triangle is
Copyright © 2007 Pearson Education, Inc. Slide 10-14
10.2 Using Heron’s Formula to Find an Area
Example The distance “as the crow flies” from Los Angeles
to New York is 2451 miles, from New York to Montreal is 331 miles, and from Montreal to Los Angeles is 2427 miles. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of the earth.)
Solution
))()(( csbsass A
5.2604)24273312451(2
1s
)24275.2604)(3315.2604)(24515.2604(5.2604
700,401
Copyright © 2007 Pearson Education, Inc. Slide 10-15
Area of a Triangle
In any triangle ABC, the area A is given by any of the following:
10.2 Area of a Triangle Given SAS
• The area of any triangle is given by A = ½bh, where b is its base and h is its height.
Achch
A sinorsin
Abcbh sin21
21
Area A
BacCabAbc sin21
and,sin21
,sin21
AAA