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6.2 Law of Cosines. What would happen if we were given:. a = 6, b = 4, C = 60 º. 6. 4. c. = . = . Sin A. Sin B. Sin 60 º. What would happen if we were given:. a = 4, b = 5, c = 3. 4. 5. 3. = . = . Sin A. Sin B. Sin C. Law of Cosines.

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6.2 Law of CosinesWhat would happen if we were given: a = 6, b = 4, C = 60º64c= = Sin ASin BSin 60ºWhat would happen if we were given: a = 4, b = 5, c = 3453= = Sin ASin BSin CLaw of CosinesUse Law of Cosines when the given information is: SSS (three lowercase letters) SAS (2 sides and an angle, all 3 letters) Law of Cosines222ab+ c – 2bc Cos A= 222ba+ c – 2ac Cos B= 222ca+ b – 2ab Cos C= Alternative FormSolve these three equations for: Cos ACos B 222222a + b - cb + c - a= Cos C= 2ab2bc222a + c - b= 2aca = 8; b = 19; c = 14 (SSS)→ Find the largest angle first 222222a + c - b8 + 14 - 19= Cos B= 2ac2(8)(14)→ remember to use the inverse Cosine to find an angle B = 116.8ºa = 8; b = 19; c = 14; B = 116.8ºOnce you know one angle, it is easiest to now use the Law of Sines. 81914= = Sin ASin 116.8ºSin C A = 22.08º C = 41.12ºSolve the following triangles:a = 5; b = 8; c = 9 a = 9; b = 7; c = 10 a = 5; b = 8; c = 9 C = 84.3º A = 33.6º B = 62.2º222222a + b - c5 + 8 - 9= Cos C= 2ab2(5)(8)59= Sin ASin 84.3ºa = 10; b = 7; c = 9 A = 76.2º B = 42.8º C = 61º222222b + c - a7 + 9 - 10= Cos A= 2bc2(7)(9)107= Sin 76.2ºSin BA = 115º; b = 15; c = 10 (SAS)What do we have enough information to solve for? 222ab+ c – 2bc Cos A= 222a= + 10 – 2(15) (10) Cos 115º152aa= = 451.7854821.3aA = 115º; b = 15; c = 10 = 21.3Now use the Law of Sines 21.31510= = Sin BSin CSin 115º B = 39.7º C = 25.2ºSolve the following triangles:a = 6; b = 4; C = 60º a = 3; c = 2; B = 110º a = 6; b = 4; C = 60ºc = 5.29B = 40.9ºA = 79.1º222ca+ b – 2ab Cos C= 45.29= Sin 60ºSin Ba = 3; c = 2; B = 110ºb = 4.14C = 27ºA = 43º222ba+ c – 2ac Cos B= 24.14= Sin 110ºSin CFind the missing information:βc5d45º82225+ 8 – 2(5) (8) Cos 45ºd= d= 5.69 Find the missing information:8β135ºc5d45º82225+ 8 – 2(5) (8) Cos 135ºc= d= 12.07 Find the missing information:35120ºc25dθ3522225+ 35 – 2(25) (35) Cos 120ºc= c= 52.2Find the missing information:8120ºc25d60ºθ3522225+ 35 – 2(25) (35) Cos 60ºd= d= 31.26.2 Law of CosinesHeron’s Area FormulaHeron’s Area FormulaAny triangle with given sides of lengths a, b, and c, has an area of: Area = s (s – a) (s – b) (s – c) where s = a + b + c2Find the area of a triangle have sides of lengths a = 43 meters, b = 53 meters, and c = 72 meters. a = 43 b = 53 c = 72s = Area = 84 (84 – 43) (84 – 53) (84 – 72)43 + 53 + 72= 842= 1131.89 square meters7 in.5 in. a = 5 b = 7 c = 10s = Area = 11 (11 – 5) (11 – 7) (11 – 10)10 in.5 + 7 + 10= 112= 16.25 square inchesA radio tower 500 feet high is located on the side of a hill with an inclination to the horizontal of 5º. How long should two guy wires be if they are to connect to the top of the tower and be secured at two points 100 feet directly above and directly below the base of the tower?1001005º

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