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Solving Triangles Using Law of Cosines

Learn how to solve triangles using Law of Cosines in trigonometry. Explore examples and calculations for finding angles and sides with given information. Understand why Law of Sines cannot be applied in certain scenarios.

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Solving Triangles Using 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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