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5.8 Law of Cosines

5.8 Law of Cosines. Law of Cosines. a 2 = b 2 + c 2 – 2bc(cosA) b 2 = a 2 + c 2 – 2ac(cosB) c 2 = a 2 + b 2 – 2ab(cosC). A=120 o , b=9, c=5 **We can’t use Law of Sines. Why?** THINK: Since we know the Angle, use the corresponding Law of Cosines: a 2 = b 2 + c 2 – 2bc(cosA)

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5.8 Law of Cosines

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  1. 5.8 Law of Cosines

  2. Law of Cosines • a2 = b2 + c2 – 2bc(cosA) • b2 = a2 + c2 – 2ac(cosB) • c2 = a2 + b2 – 2ab(cosC)

  3. A=120o, b=9, c=5 **We can’t use Law of Sines. Why?** THINK: Since we know the Angle, use the corresponding Law of Cosines: a2 = b2 + c2 – 2bc(cosA) a2 = 92 + 52 – 2(9)(5)(cos120) Calculate, then take square root. Put answer to the nearest tenth: Finish Using Law of Sines: 12.39 Sin 120 Sin B Find Last Remaining Angle: 180 – 120 – 39.3 = A 120 9 5 C B Example 1: Given 2 Sides 1 Angle = a = 12.3 B = 39.3 C = 20.7

  4. a = 24, b = 40, c = 18 THINK: Since we don’t know the Angle, use the LONGEST side Law of Cosines to Find the Angle: b2 = a2 + c2 – 2ac(cosB) 402 = 242 + 182 – 2(24) (18)(cosB) Calculate: B= cos-1(402 – 242 – 182) – 2(24)(18) Finish With Law of Sines 24 40 Sin A Sin 144.1 Find Remaining Angle: 180 – 144.1 – 20.6 = Example 2: Given All Sides A 18 40 B C 24 A = 20.6 B = 144.1 C = 15.3 =

  5. Remember Area Formula – Depends on what you have! Either 2 sides and an Included angle, or 2 angles and a side Example: a = 4, b = 7, c = 9 **Think, I know all three sides, so I can use Law of Cosines to find and angle… That will give me 2 sides and an included angle… I will use: a2 = b2 + c2 – 2bc(cosA) Calculate: A = cos-1(42 – 72 – 92) – 2(7)(9) A = 25.2o K = bc(sinA) 2 K = (7)(9)(sin25.2) 2 K = 13.4 Example 3: Finding Area

  6. Hero’s Formula! • If you know all three sides of a triangle, use Hero’s formula to find AREA. K = s = a+b+c 2 s(s-a)(s-b)(s-c) s(s-a)(s-b)(s-c) s = semi - perimeter

  7. a = 72, b = 83, c =95 First, Find the semi-perimeter – Find “s” Then Use Hero’s Formula! s = 72+83+95 2 s = 125 K = K = 2889.2 Example 4: Find Area 125(125-72)(125-83)(125-95)

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