The document discusses the Law of Cosines, which is an important extension of the Pythagorean theorem. It provides formulas for calculating sides and angles of triangles using cosine. Examples are given to demonstrate solving triangles when given side-angle-side information or all three sides. The area of triangles can be found using several formulas. Heron's formula is presented for finding the area given all three sides. Sample problems are worked through applying these concepts and formulas to solving triangles and finding areas, volumes, and distances using right triangles and trigonometry.

1. 5.6
Law of
Cosines
Copyright © 2011 Pearson, Inc.

2. What you’ll learn about
 Deriving the Law of Cosines
 Solving Triangles (SAS, SSS)
 Triangle Area and Heron’s Formula
 Applications
… and why
The Law of Cosines is an important extension of the
Pythagorean theorem, with many applications.
Copyright © 2011 Pearson, Inc. Slide 5.6 - 2

3. Law of Cosines
Let VABC be any triangle with sides and angles
labeled in the usual way. Then
a2  b2  c2  2bc cosA
b2  a2  c2  2ac cosB
c2  a2  b2  2abcosC
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4. Example Solving a Triangle (SAS)
Solve VABC given that a  10, b  4 and C  25o.
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5. Example Solving a Triangle (SAS)
Solve VABC given that a  10, b  4 and C  25o.
Use the Law of Cosines to find side c:
c2  a2  b2  2abcosC
c2  16 100  2(4)(10)cos25o
c  6.6
Use the Law of Cosines again:
102  16  43.56  2(4)(6.6)cosA
cosA  0.7659
A  140o
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6. Example Solving a Triangle (SAS)
Solve VABC given that a  10, b  4 and C  25o.
Now find (sum of the angles in a triangle = 180º):
B  180o 140o  25o  15o
The six parts of the triangle are:
A  140o a  10,
B  15o b  4,
C  25o c  6.6
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7. Area of a Triangle
1 1 1
V Area  bc sin A  ac sin B  ab sin
C
2 2 2
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8. Heron’s Formula
Let a, b, and c be the sides of VABC, and let s denote
the semiperimeter (a  b  c) / 2. Then the area of
VABC is given by
Area  ss  as  bs  c.
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9. Example Using Heron’s Formula
Find the area of a triangle with sides 10, 12, 14.
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10. Example Using Heron’s Formula
Find the area of a triangle with sides 10, 12, 14.
Compute s: s  (10 12 14) / 2  18.
Use Heron's Formula:
A  1818 1018 1218 14
= 3456
=24 6  58.8
The area is approximately 58.8 square units.
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11. Quick Review
Find an angle between 0o and 180o that is a solution
to the equation.
1. cos A  4 / 5
2. cos A  -0.25
Solve the equation (in terms of x and y) for
(a) cos A and (b) A, 0  A  180o.
3. 72  x2  y2  2xy cos A
4. y2  x2  4  4x cos A
5. Find a quadratic polynomial with real coefficients
that has no real zeros.
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12. Quick Review
Find an angle between 0o and 180o that is a solution
to the equation.
1. cos A  4 / 5 36.87o
2. cos A  0.25 104.48o
Solve the equation (in terms of x and y) for
(a) cos A and (b) A, 0  A  180o.
3. 72  x2  y2  2xy cos A
(a)
49  x2  y2
2xy
(b) cos-1 49  x2  y2
2xy

 

 
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13. Quick Review
Solve the equation (in terms of x and y) for
(a) cos A and (b) A, 0  A  180o.
4. y2  x2  4  4x cos A
(a)
y2  x2  4
4x
(b) cos-1 y2  x2  4
4x






5. Find a quadratic polynomial with real coefficients
that has no real zeros.
One answer: x2  2
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14. Chapter Test
1. Prove the identity cos3x  4cos3 x  3cos x.
2. Write the expression in terms of sin x and cos x.
cos2 2x  sin2x
3. Find the general solution without using a calculator.
2cos2x  1
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15. Chapter Test
4. Solve the equation graphically. Find all solutions
in the interval [0,2 ). sin4 x  x2  2
5. Find all solutions in the interval [0,2 ) without
using a calculator. sin2 x  2sin x  3  0
6. Solve the inequality. Use any method, but give
exact answers. 2cos x  1 for 0  x  2
7. Solve VABC, given A  79o, B  33o, and a  7.
8. Find the area of VABC, given a  3, b  5, and c  6.
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16. Chapter Test
9. A hot-air balloon is seen over Tucson, Arizona,
simultaneously by two observers at points A and B
that are 1.75 mi apart on level ground and in line
with the balloon. The angles of elevation are as
shown here. How high above ground is the balloon?
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17. Chapter Test
10. A wheel of cheese in the shape of a right circular
cylinder is 18 cm in diameter and 5 cm thick. If a
wedge of cheese with a central angle of 15º is cut
from the wheel, find the volume of the cheese
wedge.
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18. Chapter Test Solutions
1. Prove the identity cos3x  4cos3 x  3cos x.
cos3x  cos(2x  x)  cos2x cos x  sin2x sin x
 cos2 x  sin2 xcos x 2sin x cos xsin x
 cos3 x  3cos x sin2 x
 cos3 x  3cos x 1 cos2  x 4cos3 x  3cos x.
2. Write the expression in terms of sin x and cos x.
cos2 2x  sin2x 1 4sin2 x cos2 x  2cos x sin x
3. Find the general solution without using a calculator.
2cos2x  1

6
 2n ,
5
6
 2n
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19. Chapter Test Solutions
4. Solve the equation graphically. Find all solutions
in the interval [0,2 ). sin4 x  x2  2 x  1.15
5. Find all solutions in the interval [0,2 ) without
using a calculator. sin2 x  2sin x  3  0
3
2
6. Solve the inequality. Use any method, but give
exact answers. 2cos x  1 for 0  x  2

3
,

5
3

 
 
7. Solve VABC, given A  79o, B  33o, and a  7.
C  68o, b  3.88, c  6.61
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20. Chapter Test Solutions
8. Find the area of VABC, given a  3, b  5, and c  6.
 7.5
9. A hot-air balloon is seen over Tucson, Arizona,
simultaneously by two observers at points A and B
that are 1.75 mi apart on level ground and in line with
the balloon. The angles of elevation are as shown
here. How high above ground is the balloon?
≈ 0.6 mi
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21. Chapter Test Solutions
10. A wheel of cheese in the shape of a right circular
cylinder is 18 cm in diameter and 5 cm thick. If a
wedge of cheese with a central angle of 15º is cut
from the wheel, find the volume of the cheese
wedge.
405π/24 ≈ 53.01
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