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9.4 The Law of Cosines

9.4 The Law of Cosines. Objective To use the law of cosines to find unknown parts of a triangle. Data Required for Solving Oblique Triangles. Case 1 One side and two angles known: SAA or ASA Case 2 Two sides and one angle not included between the sides known: SSA

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9.4 The Law of Cosines

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  1. 9.4 The Law of Cosines Objective To use the law of cosines to find unknown parts of a triangle.

  2. Data Required for Solving Oblique Triangles Case 1 One side and two angles known: SAA or ASA Case 2 Two sides and one angle not included between the sides known: SSA  This case may lead to more than one solution. Ambiguous Case Case 3 Two sides and one angle included between the sides known: SAS Case 4 Three sides are known: SSS

  3. Solving an SAS Triangle The Law of Sines was good for ASA - two angles and the included side SAA - two angles and any side SSA - two sides and an opposite angle (being aware of possible ambiguity) Why would the Law of Sines not work for an SAS triangle? 26° 15 12.5 No side opposite from any angle to get the ratio

  4. Deriving the Law of Cosines Write an equation using Pythagorean theorem for shaded triangle. C b h a k c - k A B c

  5. Deriving the Law of Cosines Write an equation using Pythagorean theorem for shaded triangle. C h b a  - A k c A B c + k c + (–bcosA) c–bcosA

  6. Law of Cosines Similarly Note the pattern Use these to findmissing sides Opposite Include S A S

  7. Law of Cosines When C = 90o, cosC= 0, then the Law of Cosines reduces to c2 = a2 + b2, which indicates a right triangle. When C < 90o, cosC > 0, then c2 < a2 + b2 by the amount 2abcosC. When C > 90o, cosC < 0, then c2 > a2 + b2 by the amount –2abcosC.

  8. Law of Cosines If we solve the law of cosines for cos A, cos B, cos C, we obtain Use these to find missing angles S S S

  9. Choose the Method Solving an Oblique Triangle

  10. The Law of Sines and Cosines The law of sines does not distinguish between obtuse and acute angles, however, because both types of angle have positive sine values. Using the law of cosines, we can easily identify acute and obtuse angles.

  11. Applying the Cosine Law Now use it to solve the triangle we started with Label sides and angles Side c first C 26° 15 12.5 A B c Case 3 Two sides and one angle included between the sides known: • SAS

  12. Now calculate the angles use b2 = a2 + c2 – 2ac cos B and solve for B. Applying the Cosine Law C 26° 15 12.5 A B 60.25o 93.75o c = 6.65  93.75 • The remaining angledetermined by subtraction A = 180o – 93.75o – 26o = 60.25o

  13. Example 1: The leading edge of each wing of the B-2 Stealth Bomber measures 105.6 feet in length. The angle between the wing's leading edges is 109.05o. What is the wing span (the distance from A to C)? Applying the Cosine Law: Wing Span C 105.6 A [Solution 1] Not use the Law of Cosine  172 ft [Solution 2] Use the Law of Cosine  172 ft

  14. Try It Out Determine the area of these triangles 42.8° 17.9 127° 24 12 76.3° a) b) A 60.9o a 115 u2 97.9 u2

  15. Cost of a Lot Example 2: An industrial piece of real estate is priced at $4.15 per square foot. Find the cost of a triangular lot measuring 324 feet by 516 feet by 412 feet. 324 412 516 Case 4 Three sides are known: • SSS [Solution] Use the Law of Cosines to find the angle opposite to the shortest side. A

  16. Applying the Cosine Law Case 4 Three sides are known: • SSS Example 3: The lengths of the sides of a triangle are 5, 10, and 12. Solve the triangle. [Solution] Make a sketch

  17. Example 4: In the diagram at the right, AB = 5, BD = 2, DC = 4, and CA = 7. Find AD. [Solution] First we apply the law of cosines to ABC to find out the cosB: Then we apply the law of sines to ABD to find out AD: Thus AD= 5.

  18. Law of Cosines - SSA X Example 5: In XYZ, Y = 20o,z = 53, y = 19. Find YZ or x. 53 19 19 [Solution] This situation is SSA, which may have ambiguous cases. It seems we have to apply the law of sines. However, the question is only asks to find a side, YZ, or x. We then use the law of cosines. 20o Y x Z Z y2 =x2 + z2 – 2·x·z·cosY 55.5 192 =x2 + 532 – 2·x·53·cos20o 44.1 x2– 106·cos 20ox+ (532– 192) = 0

  19. Law of Cosines - SSA X Example 5: In XYZ, Y = 20o,z = 53, y = 19. Find YZ or x. 53 19 19 20o Y x Z Z 44.1 55.5

  20. 117.3 C 26.4 8 36.3 6 A B 12 Example 6: Find the three angles of the triangle. [Solution] Find the angle opposite the longest side first. Law of Sines:

  21. C 6.2 75 A B 9.5 Example 7: Solve the triangle. [Solution] 67.8 9.9 Law of Cosines: 37.2 Law of Sines:

  22. N 53 43 mi 60 67 Example 8: Application Two ships leave a port at 9 A.M. One travels at a bearing of N 53 W at 12 mph, and the other travels at a bearing of S 67 W at 16 mph. How far apart will the ships be at noon? At noon, the ships have traveled for 3 hours. 36 mi Angle C = 180 – 53 – 67 = 60 c C 48 mi The ships will be approximately 43 miles apart.

  23. Assignment P. 352 #1 – 18, 22

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