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9.6 Apply the Law of Cosines In which cases can the law of cosines be used to solve a triangle? What is Heron’s Area Formula? What is the semiperimeter?

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Presentation on theme: "9.6 Apply the Law of Cosines In which cases can the law of cosines be used to solve a triangle? What is Heron’s Area Formula? What is the semiperimeter?"— Presentation transcript:

1 9.6 Apply the Law of Cosines In which cases can the law of cosines be used to solve a triangle? What is Heron’s Area Formula? What is the semiperimeter?

2 Law of Cosines Use the law of cosines to solve triangles when two sides and the included angle are known (SAS), or when all three side are known (SSS).

3 Solve ABC with a = 11, c = 14, and B = 34°. SOLUTION Use the law of cosines to find side length b. b 2 = a 2 + c 2 – 2ac cos B b 2 = – 2(11)(14) cos 34° b Law of cosines Substitute for a, c, and B. Simplify. Take positive square root.

4 Use the law of sines to find the measure of angle A. sin A a sin B b = sin A 11 = sin 34° 7.85 sin A = 11 sin 34° A sin – ° Law of sines Substitute for a, b, and B. Multiply each side by 11 and Simplify. Use inverse sine. The third angle C of the triangle is C 180° – 34° – 51.6° = 94.4°. In ABC, b 7.85, A 51.68, and C ANSWER

5 Solve ABC with a = 12, b = 27, and c = 20. SOLUTION First find the angle opposite the longest side, AC. Use the law of cosines to solve for B. b 2 = a 2 + c 2 – 2ac cos B 27 2 = – 2(12)(20) cos B 27 2 = – 2(12)(20) = cos B – cos B B cos –1 (– ) 112.7° Law of cosines Substitute. Solve for cos B. Simplify. Use inverse cosine.

6 Now use the law of sines to find A. sin A a = sin B b sin A 12 sin 112.7° 27 = sin A= 12 sin 112.7° A sin – ° Law of sines Substitute for a, b, and B. Multiply each side by 12 and simplify. Use inverse sine. The third angle C of the triangle is C 180° – 24.2° – 112.7° = 43.1°. In ABC, A 24.2, B 112.7, and C ANSWER

7 Science Scientists can use a set of footprints to calculate an organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to 180°, the more efficiently the organism walked. The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B. SOLUTION b 2 = a 2 + c 2 – 2ac cos B = – 2(155)(197) cos B = – 2(155)(197) = cos B – cos B B cos –1 (– ) 127.3° Use inverse cosine. Simplify. Solve for cos B. Substitute. Law of cosines The step angle B is about 127.3°. ANSWER

8 Find the area of ABC. 1. a = 8, c = 10, B = 48° SOLUTION Use the law of cosines to find side length b. b 2 = a 2 + c 2 – 2ac cos B b 2 = – 2(8)(10) cos 48° b Law of cosines Substitute for a, c, and B. Simplify. Take positive square root.

9 Use the law of sines to find the measure of angle A. sin A a sin B b = sin A 8 = sin 48° 7.55 sin A = 8 sin 48° A sin – ° Law of sines Substitute for a, b, and B. Multiply each side by 8 and simplify. Use inverse sine. The third angle C of the triangle is C 180° – 48° – 51.6° = 80.4°. In ABC, b 7.55, A 51.6°, and C 80.4 °. ANSWER

10 16 2 = – 2(14)(9) cos B Find the area of ABC. 2. a = 14, b = 16, c = 9 SOLUTIONFirst find the angle opposite the longest side, AC. Use the law of cosines to solve for B. b 2 = a 2 + c 2 – 2ac cos B 16 2 = – 2(14)(9) = cos B Law of cosines Substitute. Solve for cos B. – cos B B cos –1 (– ) 85.7° Simplify. Use inverse cosine.

11 sin A a = sin B b sin A 14 sin 85.2° 16 = sin A= 14sin 85.2° Law of sines Substitute for a, b, and B. Multiply each side by 14 and simplify. Use the law of sines to find the measure of angle A. The third angle C of the triangle is C 180° – 85.2° – 60.7° = 34.1°. A sin – ° Use inverse sine. In ABC, A 60.7°, B 85.2°, and C 34.1°. ANSWER

12 Heron’s Area Formula Heron (Hero) of Alexandria, the Greek mathematician ( A.D.) is credited with using the law of cosines to find this formula for the area of a triangle.

13 Urban Planning The intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown. SOLUTIONSTEP 1 Find the semiperimeter s. s = (a + b + c ) = ( ) = 380 STEP 2 Use Heron’s formula to find the area of ABC. Area = s (s – a) (s – b) (s – c) 380 (380 – 170) (380 – 240) (380 – 350)= 18,300 The area of the traffic triangle is about 18,300 square yards.

14 Find the area of ABC. 4. SOLUTION STEP 1 Find the semiperimeter s. s = (a + b + c ) = ( ) = 12 STEP 2 Use Heron’s formula to find the area of ABC. Area = s (s – a) (s – b) (s – c) The area is about 18.3 square units. 12 (12 – 8) (12 – 11) (12 – 5)= 18.3

15 Find the area of ABC. 5. SOLUTION STEP 1 Find the semiperimeter s. s = (a + b + c ) = ( ) = 10 STEP 2 Use Heron’s formula to find the area of ABC. Area = s (s – a) (s – b) (s – c) The area is about 13.4 square units. 10 (10– 4) (10 – 9) (10 – 7)= 13.4

16 Find the area of ABC. 6. SOLUTION STEP 1 Find the semiperimeter s. s = (a + b + c ) = ( ) = 25 STEP 2 Use Heron’s formula to find the area of ABC. Area = s (s – a) (s – b) (s – c) The area is about 80.6 square units. 25 (25– 15) (25 – 23) (25 – 12)= 80.6

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18 9.6 Assignment Page 596, 3-39 every 3 rd problem No work is the same as a missing problem


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