Unit 5.6

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.
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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
Copyright © 2011 Pearson, Inc. Slide 5.6 - 20

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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