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6.6 Adjusting the Pythagorean Theorem. C. c 2 = a 2 + b 2. b. a. The Cosine Law. c. B. A. 5. 3. 4. We start with a rope 12 units long. Units are marked. We form the following figure. A right angle is formed. 3, 4, 5 is a Pythagorean Triple.
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6.6 Adjusting the Pythagorean Theorem C c2 = a2 + b2 b a The Cosine Law c B A
5 3 4 We start with a rope 12 units long. Units are marked. We form the following figure. A right angle is formed. 3, 4, 5 is a Pythagorean Triple
A Try the Pythagorean TheoremÐC < 90º 51º 6.9 cm 4.7 cm Compare c2 with a2 + b2 43º c2a2 + b2 86º B C 5.3 cm 6.92 5.32 + 4.72 c2 < a2 + b2
Try the Pythagorean Theorem ÐC > 90º A Compare c2 with a2 + b2 46º c2a2 + b2 7.4 cm 7.42 5.32 + 4.72 4.7 cm c2 > a2 + b2 39º 95º B C 5.3 cm
The Cosine Law C b a c2 = a2 + b2 – 2ab cos C a2 = b2 + c2 – 2bc cos A c B A b2 = a2 + c2 – 2ac cos B The Cosine Law is used for situations involving SAS as well as SSS. You are given 2 sides and the contained angle and you wish to find the third side or three side and you need to find one of the angles.
Example (1) Find a. a2 = b2 + c2 – 2bc cos A a2 = 4.62 + 6.22 – 2(4.6)(6.2) cos 52º A a2 = 21.16 + 38.44 – 35.12 52º a2 = 24.48 6.2 cm 4.6 cm a= 4.9 cm a B C
b = Example (2) Find b. b2 = a2 + c2 – 2ac cos B b2 = 5.02 + 5.72 – 2(5.0)(5.7) cos 78º b2 = 25 + 32.49 – 11.85 A b2 = 45.64 b 5.7 cm 78º b = 6.8 B C 5.0 cm
– 96 = cos P – 144 R Example (3) Find ÐP. p2 = q2 + r2 – 2qrcos P 8 7 72 = 82 + 92 – 2(8)(9) cos P Q 49 = 64 + 81– 144 cos P P 9 49 – 64 – 81 = – 144 cos P ÐP = 48.2º – 96 = – 144 cos P 0.6666666 = cos P ÐP = cos-1(0.66666)
Example 4 Two girls begin cycling from the same location. The angle of the roads is 41º. One girl is cycling at 14 km/h and the second girl is cycling at 16 km/h. How far apart are the girls after 3 hours? 42 km 14 km/h x 3 hours 41º 41º 16 km/h 48 km x2 = 422 + 482 – 2(42)(48)cos 41º x2 = 1025
A b c In DADC h a – x x B D C a In DABD Deriving the Cosine Law: Given DABC and all angles are acute. Draw altitude AD.
A ÐA = cos-1(0.0370) Given: SSS (finding the angle) Ex:Find the largest angle 15 18 Find Ð A. a2 = b2 + c2 – 2bccos A B C 23 232 = 182 + 152 – 2(18)(15) cos A 529 = 324 + 225– 540 cos A 540 cos A = 324 + 225– 529 cos A = 0.0370 ÐA = 87.9°
b= B Ex: If a = 39 cm, Ð B = 48° and c = 57 cm, 48º 57 cm Find b. 39 cm b2 = a2 + c2 – 2ac cos B A C b b2 = 392 + 572 – 2(39)(57) cos 48º b2 = 1521 + 3249– 2975 b2 = 1794 b = 42.4 cm