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1-6 Applying the Cosine Law

Chapter 1 - Trigonometry. 1-6 Applying the Cosine Law. Solving an SAS Triangle. The Law of Sines was good for ASA - two angles and the included side AAS - two angles and any side SSA - two sides and an opposite angle Why would the Law of Sines not work for an SAS triangle?. 15. 26°.

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1-6 Applying the Cosine Law

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  1. Chapter 1 - Trigonometry 1-6 Applying the Cosine Law

  2. Solving an SAS Triangle • The Law of Sines was good for • ASA - two angles and the included side • AAS - two angles and any side • SSA - two sides and an opposite angle • Why would the Law of Sines not work for an SAS triangle? 15 26° No side opposite from any angle to get the ratio 12.5

  3. Deriving the Law of Cosines C Some acute triangles that cannot be solved using the sine law can be solved using the cosine law. b h a x c - x B A D In ΔABC, draw CD perpendicular to AB. c CD is the altitude, h, of ΔABC Let AD = x, BD = c - x In ΔACD, b2 = h2 + x2

  4. Deriving the Law of Cosines C In ΔBCD, a2 = h2 + (c – x)2 b h a By expanding: a2 = h2 + c2 – 2cx + x2 By rearranging: a2 = c2 + (h2 + x2) – 2cx x c - x B A D c Substituting b2 for (h2 + x2) and b cos A for x gives: a2 = c2 + b2 – 2c(b cos A) a2 = c2 + b2 – 2cb cos A a2 = b2 + c2 – 2bc cos A

  5. Deriving the Law of Cosines C Deriving similar equations that include cos B and cos C gives three forms of the cosine law. b h a x c - x B A D c a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C

  6. Applying the Cosine Law • Now use it to solve the triangle we started with • Label sidesand angles • Side c first C 26° 15 12.5 A B c c2 = a2 + b2 – 2ab cos C c2 = 12.52 + 152 – 2(12.5)(15) cos 26o

  7. Applying the Cosine Law C 26° 15 12.5 A B c = 6.65 Now calculate the remaining angles:

  8. Applying the Cosine Law • The remaining angledetermined by subtraction • 180 – 93.75 – 26 = 60.25 C 26° 15 12.5 A B c = 6.65

  9. Example 1 • Solve for angle A, B, and C using the Cosine Law. • Solve for angle A. a2 = b² + c² - 2bcCosA 16.5² = 10² + 15² - 2(10)(15)CosA 16.5² - 10² - 15² = CosA -2(10)(15) Cos-1 A = 16.5² - 10² - 15² -2(10)(15) Angle A = 80°

  10. Example 1 continued b) Solve for Angle B. b2 = a² + c² - 2ac Cos B 10² = 16.5² +15² - 2(16.5)(15)Cos B 10² - 16.5² - 15² = Cos B -2(16.5)(15) Cos-1B = 10² - 16.5² - 15² -2(16.5)(15) Angle B = 37ᵒ

  11. Example 1 continued c) Solve for Angle C c2 = a² + b² - 2abCosC 15² = 16.5² +10² - 2(16.5)(10)Cos C 15² - 16.5² - 10² = Cos C -2 (16.5)(10) Cos-1C = 15² - 16.5² - 10² -2(16.5)(10) Angle C = 63ᵒ

  12. Example 2 Solve for the length of side f. f2 = d² + e² - 2deCos F f² = 15² + 12² - 2(15)(12)Cos 36° f² = 78 f = 9m

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