Unit 5 Day 5 Law of Cosines

Happiness begins where selfishness ends. - John Wooden Warm-up Happiness begins where selfishness ends. - John Wooden x = 2.25 x = -5 Solve each triangle using Law of Sines. 3) 4) C = 40 a = 32.97 b = 13.99 A = 83 a = 10 c = 8.05

HW Packet p. 9 odds 7. Two Solutions Case-1 Case-2 43.7 136.3 43.7 136.3 102.3  9.7  148.7 units 25.6 units 9. One Solution 20.2 34.8 62.7 units Questions 11 & 13 Require Law of Cosines to solve (SAS).

HW Packet p. 9 evens 6. One Solution 54 63  44 units 8. One Solution 23 129 248.3 units 10. Two Solutions Case-1 Case-2 53.8 126.2 95.2  20.8  49.4 units 17.6 units 12. Case-1 Case-2 17.5 162.5 146.5  1.5  22.03 units 1.04 units Questions 11 & 13 Require Law of Cosines to solve (SAS).

Law of Cosines

a2 = b2 + c2 – 2bcCosA a2 = 102 + 82 – 2(10)(8)Cos(60) Ex. 1 Use Law of Cosines to find a. Notice this arrangement is SAS. a2 = b2 + c2 – 2bcCosA a2 = 102 + 82 – 2(10)(8)Cos(60) a2 = 164 – 160Cos(60)

r2 = s2 + q2 – 2sqCosR 232 = 372 + 182 – 2(37)(18)Cos(R) Ex. 2 Use Law of Cosines to find mR. Notice this arrangement is SSS. r2 = s2 + q2 – 2sqCosR 232 = 372 + 182 – 2(37)(18)Cos(R) 529 = 1693 – 1332Cos(R) R =cos-1(-1164/1332) R = 29.1

Ex. 3 Tips for steps: You’ll need 3 steps just as you did with Law of Sines Problems Think BIG – use law of Cosines with CAPITAL letter (given angle) first Short Side Second with Law of Sines Subtract angles from 180° k2 = 142 + 182 – 2(14)(18)Cos(51) Twalden: These Steps are critical for Law of Cosines IF you are doing Law of Sines for step 2. Chelsea Mauldin tells me that if the students use Law of Cosines for all the steps, the problems work fine too. Sin-1 L = 180 – 51 – 49.81 = 79.19

Extra Example – not in notes - Solving given SSS Solve triangle ABC if a=8, b=10, and c=5. Remember, if no picture is given, draw one!  Tips for steps: You’ll need 3 steps just as you did with Law of Sines Problems Think BIG – use law of Cosines with BIGGEST side first Short Side Second with Law of Sines Subtract angles from 180° 102 = 52 + 82 – 2(5)(8)Cos(B) Twalden: The essential step here is Step 1..big first. Short Side Second is not essential here BUT it is for the other case…so I usually get my AFM students to do short side 2nd for both so they don’t have to remember which it’s essential for. Chelsea Mauldin tells me that if the students use Law of Cosines for all the steps, the problems work fine too. Sin-1 L = 180 – 97.9 – 29.67 = 52.43

m<A= 34o , m<B= 108o , m<C= 38o Ex. 4, 5 YOU TRY Solve each triangle. a = 33 , m<B= 31o , m<C= 58 m<A= 34o , m<B= 108o , m<C= 38o

What type of triangle do you have? Ambiguous Case Law of Sines Which Formula Do I Use? What type of triangle do you have? RIGHT OBLIQUE Pythagorean Theorem & Trig Ratios AAS,ASA SSS,SAS SSA Ambiguous Case Law of Sines Law of Sines Law of Cosines

Practice!

Capital letters are angles, lowercase letters are sides Practice Answers! Capital letters are angles, lowercase letters are sides B = 86 C = 56 D = 38 H = 31 G = 109 g = 14.7 p = 6.9 M = 79 Q = 63 22.) B = 61, b = 5.4, a = 4.1 23.) c = 6.3, A = 80, B = 63 24.) A = 30, B = 39, C = 111 25.) B = 99, b = 31.3, a = 25.3