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5.6 
Law of 
Cosines 
Copyright © 2011 Pearson, Inc.
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
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 3
Example Solving a Triangle (SAS) 
Solve VABC given that a  10, b  4 and C  25o. 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 4
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 5
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 6
Area of a Triangle 
1 1 1 
V Area  bc sin A  ac sin B  ab sin 
C 
2 2 2 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 7
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. 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 8
Example Using Heron’s Formula 
Find the area of a triangle with sides 10, 12, 14. 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 9
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. 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 10
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. 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 11
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 
 
  
 
  
Copyright © 2011 Pearson, Inc. Slide 5.6 - 12
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 13
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 14
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. 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 15
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? 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 16
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. 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 17
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 18
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 19
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
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 
Copyright © 2011 Pearson, Inc. Slide 5.6 - 21

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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. 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 3
  • 4. Example Solving a Triangle (SAS) Solve VABC given that a  10, b  4 and C  25o. Copyright © 2011 Pearson, Inc. Slide 5.6 - 4
  • 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 5
  • 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 6
  • 7. Area of a Triangle 1 1 1 V Area  bc sin A  ac sin B  ab sin C 2 2 2 Copyright © 2011 Pearson, Inc. Slide 5.6 - 7
  • 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. Copyright © 2011 Pearson, Inc. Slide 5.6 - 8
  • 9. Example Using Heron’s Formula Find the area of a triangle with sides 10, 12, 14. Copyright © 2011 Pearson, Inc. Slide 5.6 - 9
  • 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. Copyright © 2011 Pearson, Inc. Slide 5.6 - 10
  • 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. Copyright © 2011 Pearson, Inc. Slide 5.6 - 11
  • 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       Copyright © 2011 Pearson, Inc. Slide 5.6 - 12
  • 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 13
  • 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 14
  • 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. Copyright © 2011 Pearson, Inc. Slide 5.6 - 15
  • 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? Copyright © 2011 Pearson, Inc. Slide 5.6 - 16
  • 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. Copyright © 2011 Pearson, Inc. Slide 5.6 - 17
  • 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 18
  • 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 19
  • 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 Copyright © 2011 Pearson, Inc. Slide 5.6 - 21