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Law of Cosines 10.5

Law of Cosines 10.5. The Law of Sines cannot be used to solve every triangle. Instead, you can apply the Law of Cosines. You can use the Law of Cosines to solve a triangle if you are given • two side lengths and the included angle measure (SAS) or • three side lengths (SSS). Law of Cosines.

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Law of Cosines 10.5

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  1. Law of Cosines 10.5

  2. The Law of Sines cannot be used to solve every triangle. Instead, you can apply the Law of Cosines. You can use the Law of Cosines to solve a triangle if you are given • two side lengths and the included angle measure (SAS) or • three side lengths (SSS).

  3. Law of Cosines Standard Form Alternative Form

  4. Helpful Hint Helpful Hint The angle referenced in the Law of Cosines is across the equal sign from its corresponding side. Do not round your answer until the final step of the computation. If a problem has multiple steps, store the calculated answers to each part in your calculator.

  5. Law of Cosines – SAS Find the measure of XZ. XZ2 = XY2 + YZ2 – 2(XY)(YZ)cos Y Law of Cosines XZ2= 352 + 302 – 2(35)(30)cos 110° Substitution. Simplify. XZ2 2843.2423 Take the square root of both sides. XZ 53.3

  6. Law of Cosines – SSS Find mT. RS2 = RT2 + ST2 – 2(RT)(ST)cos T Law of Cosines 72 = 132 + 112 – 2(13)(11)cos T Substitution. Simplify. 49 = 290 – 286 cosT Subtract 290 from both sides. –241 = –286 cosT Division. Use inverse cosine to find mT.

  7. Law of Cosines – SAS Find the measure of DE. Law of Cosines DE2 = EF2 + DF2 – 2(EF)(DF)cos F DE2= 182 + 162 – 2(18)(16)cos 21° Substitution. DE2 42.2577 Simplify. Take the square root of both sides. DE 6.5

  8. Law of Cosines – SSS Find mK. JL2 = LK2 + KJ2 – 2(LK)(KJ)cos K Law of Cosines 82 = 152 + 102 – 2(15)(10)cos K Substitution. Simplify. 64 = 325 – 300 cosK Subtract 325 both sides. –261 = –300 cosK Division. Use inverse cosine to find mK.

  9. Law of Cosines – SAS Find the measure of YZ. YZ2 = XY2 + XZ2 – 2(XY)(XZ)cos X Law of Cosines Substitution. YZ2= 102 + 42 – 2(10)(4)cos 34° Simplify. YZ2 49.6770 Take the square root of both sides. YZ 7.0

  10. Law of Cosines – SSS Find mR. PQ2 = PR2 + RQ2 – 2(PR)(RQ)cos R Law of Cosines 9.62 = 5.92 + 10.52 – 2(5.9)(10.5)cos R Substitution. 92.16 = 145.06 – 123.9cosR Simplify. Subtract 145.06 both sides. –52.9 = –123.9 cosR Solve for cosR. Use inverse cosine to find mR.

  11. Law of Cosines – SAS Solve the following triangle. Law of Cosines BC2 = AB2 + AC2 – 2(AB)(AC)cos A BC2= 3.92 + 3.12 – 2(3.9)(3.1)cos 45° Substitution. Simplify. BC2 7.7222 Take the square root of both sides. BC 2.8 mi Law of Sines Substitution. Multiply both sides by 3.9. Use inverse sine to find mC.

  12. 31 m Law of Cosines – SAS An engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree. Law of Cosines AC2 = AB2 + BC2 – 2(AB)(BC)cos B Substitution. AC2= 312 + 562 – 2(31)(56)cos 100° Simplify. AC2 4699.9065 Take the square root of both sides. AC 68.6 m

  13. 31 m Find the measure of the angle the cable would make with the ground. Law of Sines Substitution. Multiply both sides by 56. Use inverse sine to find mA.

  14. Use ΔABC for a – c. Round lengths to the nearest tenth and angle measures to the nearest degree. a. mB = 20°, mC = 31° and b = 210. Find a. b. a = 16, b = 10, and mC = 110°. Find c. c.a = 20, b = 15, and c = 8.3. Find mA. 477.2 21.6 115°

  15. An observer in tower A sees a fire 1554 ft away at an angle of depression of 28°. To the nearest foot, how far is the fire from an observer in tower B? To the nearest degree, what is the angle of depression to the fire from tower B? 1212 ft; 37°

  16. Assignment Geometry: 10.1 Law of Cosines

  17. Definition: Heron’s Area Formula 10 8 5 Heron’s Area Formula Given any triangle with sides of lengths a, b, and c, the area of the triangle is given by Find the area of the triangle.

  18. The intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown. Area = 18,300 = 380 (380 – 170) (380 – 240) (380 – 350) s (s – a)(s – b)(s – c) Area 1 1 s = (a + b + c ) = (170 + 240 + 350) 2 2 EXAMPLE 4 Find the semiperimeter s. = 380 The area of the traffic triangle is about 18,300 square yds.

  19. Use Heron’s formula to find the area of ABC. Find the area of ABC. Area = 18.3 = 12 (12 – 8) (12 – 11) (12 – 5) s (s – a)(s – b)(s – c) Find the semiperimeter s. Area 1 1 s = (a + b + c ) = (5 + 8 + 11) 2 2 GUIDED PRACTICE = 12 The area is about 18.3 square units.

  20. Use Heron’s formula to find the area of ABC. Find the area of ABC. Area = 13.4 = 10 (10– 4) (10 – 9) (10 – 7) s (s – a)(s – b)(s – c) Find the semiperimeter s. Area 1 1 2 2 s = (a + b + c ) = (4 + 9+ 7) GUIDED PRACTICE = 10 The area is about 13.4 square units.

  21. Use Heron’s formula to find the area of ABC. Find the area of ABC. Area = 80.6 = 25 (25– 15) (25 – 23) (25 – 12) s (s – a)(s – b)(s – c) Find the semiperimeter s. Area 1 1 2 2 s = (a + b + c ) = (15 + 23+ 127) = 25 The area is about 80.6 square units.

  22. Assignment Geometry: 10.5 Heron’s Law

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