Grade 9 Mathematics Module 7 Answer Key

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Module 7 triangle trigonometry super final

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Module 7 triangle trigonometry super final

1. 1. GRADE 9 MATHEMATICS QUARTER 4 Module 7 May 2014
2. 2. Triangle Trigonometry MODULE 7 May 20, 2014
3. 3. In a right triangle, one of the angles measures 90 𝑜 , and the remaining two angles are acute and complementary. The longest side of a right triangle is known as the hypotenuse and is opposite the right angle. The other two sides are called legs. The leg that is a side of an acute angle is called the side adjacent to the angle. The other leg is the side opposite the angle. Lesson 1 The Six Trigonometric Ratios: Sine, Cosine, Tangent, Cosecant, Secant, and Cotangent C B A a c b Hypotenuse Side Opposite Side Adjacent 3
4. 4. Lesson 1 The Six Trigonometric Ratios: Sine, Cosine, Tangent, Cosecant, Secant, and Cotangent If two angles of a triangle are congruent to two angles of another triangle, the triangles are similar. If an acute angle of one right triangle is congruent to an acute angle of another right triangle, the triangles are similar, and the ratios of the corresponding sides are equal. Therefore, any two congruent angles of different right triangles will have equal ratios associated with them. The ratios of the sides of the right triangles can be used to define the trigonometric ratios. The ratio of the side opposite 𝜃 and the hypotenuse is known as the sine. The ratio of the side adjacent 𝜃 and the hypotenuse is known as the cosine. The ratio of the side opposite 𝜃 and the side adjacent 𝜃 is known as the tangent. 4
5. 5. SOH-CAH-TOA is a mnemonic device commonly used for remembering these ratios. sin 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 tan 𝜃 = 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 5 Lesson 1 The Six Trigonometric Ratios: Sine, Cosine, Tangent, Cosecant, Secant, and Cotangent Words Symbol Definition Trigonometric Ratios sine 𝜃 sin 𝜃 sin 𝜃 = 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cosine 𝜃 cos 𝜃 cos 𝜃 = 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 tangent 𝜃 tan 𝜃 tan 𝜃 = 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 C B A a c b Hypotenuse Side Opposite Side Adjacent
6. 6. 6 Lesson 1 The Six Trigonometric Ratios: Sine, Cosine, Tangent, Cosecant, Secant, and Cotangent Words Symbol Definition Reciprocal Trigonometric Ratios cosecant 𝜃 csc 𝜃 𝑐𝑠𝑐 𝜃 = 1 sin 𝜃 𝑜𝑟 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 secant 𝜃 sec 𝜃 sec 𝜃 = 1 cos 𝜃 𝑜𝑟 ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 cotangent 𝜃 cot 𝜃 cot 𝜃 = 1 tan 𝜃 𝑜𝑟 𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 C B A a c b Hypotenuse Side Opposite Side Adjacent These definitions are called the reciprocal identities. CHO-SHA-CAO is a mnemonic device commonly used for remembering these ratios. csc 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 sec 𝜃 = ℎ𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 cot 𝜃 = 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 In addition to the trigonometric ratios sine, cosine, and tangent, there are three other trigonometric ratios called cosecant, secant, and cotangent. These ratios are the reciprocals of sine, cosine, and tangent, respectively.
7. 7. 7 Lesson 2 Trigonometric Ratios of Special Angles Consider the special relationships among the sides of 30 𝑜 - 60 𝑜 - 90 𝑜 and 45 𝑜 - 45 𝑜 - 90 𝑜 triangles. x x 3 2x 60 𝑜 30 𝑜 y y y 2 45 𝑜 45 𝑜 These special relationships can be used to determine the trigonometric ratios for 30 𝑜 , 45 𝑜 , and 60 𝑜 .
8. 8. 𝜽 sin 𝜽 cos 𝜽 tan 𝜽 csc 𝜽 sec 𝜽 cot 𝜽 30 𝑜 45 𝑜 60 𝑜 Activity Complete the table below that summarizes the values of the trigonometric ratios of the angles 30 𝑜, 45 𝑜, and 60 𝑜. 8 Lesson 2 Trigonometric Ratios of Special Angles x x 3 2x 60 𝑜 30 𝑜 y y y 2 45 𝑜 45 𝑜
9. 9. 9 Answers: Lesson 2 Trigonometric Ratios of Special Angles 𝜽 sin 𝜽 cos 𝜽 tan 𝜽 csc 𝜽 sec 𝜽 cot 𝜽 30 𝑜 1 2 3 2 3 3 2 2 3 3 3 45 𝑜 2 2 2 2 1 2 2 1 60 𝑜 3 2 1 2 3 2 3 3 2 3 3
10. 10. 10 Lesson 2 Trigonometric Ratios of Special Angles 𝜽 sin 𝜽 cos 𝜽 tan 𝜽 csc 𝜽 sec 𝜽 cot 𝜽 30 𝑜 1 2 3 2 3 3 2 2 3 3 3 45 𝑜 2 2 2 2 1 2 2 1 60 𝑜 3 2 1 2 3 2 3 3 2 3 3 Notice that sin 30 𝑜 = cos 60 𝑜 and cos 30 𝑜 = sin 60 𝑜 . This is an example showing that the sine and cosine are cofunctions. That is, if 𝜃 is an acute angle, sin 𝜃= cos (90 𝑜 −𝜃). Similar relationships hold true for the other trigonometric ratios.
11. 11. 11 Lesson 2 Trigonometric Ratios of Special Angles 𝜽 sin 𝜽 cos 𝜽 tan 𝜽 csc 𝜽 sec 𝜽 cot 𝜽 30 𝑜 1 2 3 2 3 3 2 2 3 3 3 45 𝑜 2 2 2 2 1 2 2 1 60 𝑜 3 2 1 2 3 2 3 3 2 3 3 sin 𝜃= cos (90 𝑜−𝜃) cos 𝜃= s𝐢𝐧 (90 𝑜−𝜃) Cofunctions tan 𝜃= cot (90 𝑜−𝜃) cot 𝜃= tan (90 𝑜−𝜃) sec 𝜃= csc (90 𝑜 −𝜃) csc 𝜃= sec (90 𝑜 −𝜃)
12. 12. There are many applications that require trigonometric solutions. For example, surveyors use special instruments to find the measures of angles of elevation and angles of depression. 12 Lesson 3 Angles of Elevation and Angles of Depression
13. 13. 13 Lesson 3 Angles of Elevation and Angles of Depression An angle of elevation is the angle between a horizontal line and the line of sight from an observer to an object at a higher level.
14. 14. 14 Lesson 3 Angles of Elevation and Angles of Depression An angle of depression is the angle between a horizontal line and the line of sight from the observer to an object at a lower level.
15. 15. 15 Lesson 3 Angles of Elevation and Angles of Depression The angle of elevation and the angle of depression are equal in measure because they are alternate interior angles.
16. 16. 16 Lesson 4 Word Problems Involving Right Triangles Trigonometric functions can be used to solve word problems involving right triangles. The most common functions used are the sine, cosine, and tangent. Moreover, you can use trigonometric functions and inverse relations to solve right triangles. To solve a triangle means to find all the measures of its sides and angles. Usually, two measures are given. Then you can find the remaining measures.
17. 17. 17 Lesson 4 Word Problems Involving Right Triangles Example 1. A ladder is 12 feet long. a) If the ladder is placed against a wall so that its base is 2 feet from the wall, find, to the nearest degree, the acute angle the ladder makes with the ground. b) Suppose the base of the ladder is 𝑥 feet from the wall. Find an expression for 𝜃 , the angle the ladder makes with the ground.
18. 18. 18 Lesson 4 Word Problems Involving Right Triangles Example 1. A ladder is 12 feet long. a) If the ladder is placed against a wall so that its base is 2 feet from the wall, find, to the nearest degree, the acute angle the ladder makes with the ground. Solution. cos 𝜃 = 2 12 𝜃 = 𝑐𝑜𝑠−1 1 6 𝜃 ≈ 80 𝑜
19. 19. 19 Lesson 4 Word Problems Involving Right Triangles Example 1. A ladder is 12 feet long. b) Suppose the base of the ladder is 𝑥 feet from the wall. Find an expression for 𝜃, the angle the ladder makes with the ground. Solution. cos 𝜃 = 𝑥 12 𝜃 = 𝑐𝑜𝑠−1 𝑥 12
20. 20. 20 Lesson 4 Word Problems Involving Right Triangles Example 2. Latashi and Markashi are flying kites on a windy day. Latashi has released 250 feet of string, and Markashi has released 225 feet of string. The angle that Latashi's kite string makes with the horizontal is 35 𝑜 . The angle that Markashi's kite string makes with the horizontal is 42 𝑜 . Which kite is higher and by how much?
21. 21. 21 Lesson 4 Word Problems Involving Right Triangles Solution. For Latashi's kite: sin 35 𝑜 = ℎ 250 𝑓𝑡 ℎ = 250 𝑓𝑡 sin 35 𝑜 ℎ = 143.3941091 𝑓𝑡 Latashi's kite has a height about 143.39 ft. 250 ft 35 𝑜 Height = ?
22. 22. 22 Lesson 4 Word Problems Involving Right Triangles Solution. For Markashi's kite: sin 42 𝑜 = ℎ 225 𝑓𝑡 ℎ = 225 𝑓𝑡 sin 42 𝑜 ℎ = 150.5543864 𝑓𝑡 Markashi's kite has a height about 150.55 ft. 225 ft 42 𝑜 Height = ?
23. 23. 23 Lesson 4 Word Problems Involving Right Triangles Solution. Let's subtract the height of Markashi's kite and the height of Latashi's kite. 150.5543864 𝑓𝑡 − 143.3941091 𝑓𝑡 = 7.160277343 𝑓𝑡 Markashi's kite is higher than Latashi's kite by about 7.16 ft.
24. 24. 24 Lesson 5 Oblique Triangles Trigonometry enables sides and angle measures to be found in triangles other than right triangles. An oblique triangle is one that does not contain a right angle. Oblique triangles may be classified into two---acute and obtuse. An acute triangle is one that has three acute angles. An obtuse triangle is one that has one obtuse angle.
25. 25. 25 Lesson 5 Oblique Triangles Activity Identify the acute and obtuse triangles.
26. 26. 26 Lesson 5 Oblique Triangles Activity Identify the acute and obtuse triangles.
27. 27. 27 Lesson 5.1 The Law of Sines and Its Applications Law of Sines Let ∆𝐴𝐵𝐶 be any triangle with 𝑎, 𝑏, and 𝑐 representing the measures of the sides opposite the angles with measures 𝐴, 𝐵, and 𝐶, respectively. Then, the following are true. sin 𝐴 𝑎 = sin 𝐵 𝑏 = sin 𝐶 𝑐 𝑎 sin 𝐴 = 𝑏 sin 𝐵 = 𝑐 sin 𝐶
28. 28. 28 Lesson 5.1 The Law of Sines and Its Applications From geometry, you know that a unique triangle can be formed if you know a) the measures of two angles and the included side (ASA) or b) the measures of two angles and the non-included side (AAS). Therefore, there is one unique solution when you use the Law of Sines to solve a triangle given the measures two angles and one side.
29. 29. 29 Lesson 5.1 The Law of Sines and Its Applications From geometry, you know that c) the measures of two sides and a non-included angle (SSA) do not necessarily define a unique triangle. However, one of the following will be true. 1. No triangle exists. 2. Exactly one triangle exists. 3. Two triangles exist. In other words, there may be no solution, one solution, or two solutions. A situation with two solutions is called the ambiguous case.
30. 30. 30 Lesson 5.1 The Law of Sines and Its Applications Suppose you know the measures 𝑎 , 𝑏 , and 𝐴 . Consider the following cases.
31. 31. 31 Lesson 5.1 The Law of Sines and Its Applications ASA Case Example 1. Cartography To draw a map, a cartographer needed to find the distances between point 𝑍 across the lake and each of point 𝑋 and 𝑌 on another side. The cartographer found 𝑋𝑌 ≈ 0.3 miles, 𝑚∠𝑋 ≈ 50 𝑜 , and 𝑚∠𝑌 ≈ 100 𝑜 . Find the distances from 𝑋 to 𝑍 and from 𝑌 to 𝑍. 𝑋 𝑍 𝑌
32. 32. 32 Lesson 5.1 The Law of Sines and Its Applications ASA Example 1 Solution. 𝑚∠𝑍 ≈ 180 𝑜 − 100 𝑜 − 50 𝑜 ≈ 30 𝑜 sin 100 𝑜 𝑋𝑍 = sin 30 𝑜 0.3 𝑋𝑍 ∙ sin 30 𝑜 = 0.3 ∙ sin 100 𝑜 𝑋𝑍 = 0.3 sin 100 𝑜 sin 30 𝑜 𝑋𝑍 ≈ 0.59 The distance from 𝑋 to 𝑍 is about 0.59 miles. 𝑋 𝑍 𝑌 0.3 mi 50 𝑜 100 𝑜
33. 33. 33 Lesson 5.1 The Law of Sines and Its Applications ASA Example 1 (Continuation) Solution. 𝑚∠𝑍 ≈ 180 𝑜 − 100 𝑜 − 50 𝑜 ≈ 30 𝑜 sin 50 𝑜 𝑌𝑍 = sin 30 𝑜 0.3 Y𝑍 ∙ sin 30 𝑜 = 0.3 ∙ sin 50 𝑜 Y𝑍 = 0.3 sin 50 𝑜 sin 30 𝑜 Y𝑍 ≈ 0.46 The distance from 𝑌 to 𝑍 is about 0.46 miles. 𝑋 𝑍 𝑌 0.3 mi 50 𝑜 100 𝑜
34. 34. 34 Lesson 5.1 The Law of Sines and Its Applications AAS Case Example 2. A hill slopes upward at an angle of 5 𝑜 with the horizontal. A tree grows vertically on the hill. When the angle of elevation of the sun is 30 𝑜 , the tree casts a shadow downhill that is 32 meters long. If the shadow is entirely on the hill, how tall is the tree?
35. 35. 35 Lesson 5.1 The Law of Sines and Its Applications AAS Case (Illustration) Example 2. A hill slopes upward at an angle of 5 𝑜 with the horizontal. A tree grows vertically on the hill. When the angle of elevation of the sun is 30 𝑜 , the tree casts a shadow downhill that is 32 meters long. If the shadow is entirely on the hill, how tall is the tree?
36. 36. 36 Lesson 5.1 The Law of Sines and Its Applications AAS Example 2 Solution. a) Right Triangle Involving the Hill Alone 𝑚∠3𝑟𝑑 = 180 𝑜 − 90 𝑜 − 5 𝑜 = 85 𝑜 b) Right Triangle Involving the Tree 𝑚∠3𝑟𝑑 = 180 𝑜 − 90 𝑜 − 30 𝑜 = 60 𝑜 c) Straight Angle Involving the Hill and Tree 180 𝑜 − 85 𝑜 = 95 𝑜 d) Oblique Triangle Above the Hill 𝑚∠3𝑟𝑑 = 180 𝑜 − 95 𝑜 − 60 𝑜 = 25 𝑜
37. 37. 37 Lesson 5.1 The Law of Sines and Its Applications AAS Example 2 (Continuation) Solution. sin 25 𝑜 𝑡 = sin 60 𝑜 32 t ∙ sin 60 𝑜 = 32 ∙ sin 25 𝑜 t = 32 sin 25 𝑜 sin 60 𝑜 t ≈ 15.6 The tree is about 15.6 meters tall.
38. 38. 38 Lesson 5.1 The Law of Sines and Its Applications SSA Case Let's have to cases. a) Case 1: 𝐴 < 90 𝑜 b) Case 2: 𝐴 ≥ 90 𝑜 Example 3. Determine the number of possible solutions for each triangle. a) 𝐴 = 30 𝑜 , 𝑎 = 8, 𝑏 = 10 b) 𝑏 = 8, 𝑐 = 10, 𝐵 = 118 𝑜
39. 39. 39 Lesson 5.1 The Law of Sines and Its Applications SSA Case Example 3. Determine the number of possible solutions for each triangle. a) 𝐴 = 30 𝑜 , 𝑎 = 8, 𝑏 = 10 Since 30 𝑜 < 90 𝑜 , consider Case 1. 𝑏 𝑠𝑖𝑛 𝐴 = 10 sin 30 𝑜 𝑏 𝑠𝑖𝑛 𝐴 = 10 0.5 𝑏 𝑠𝑖𝑛 𝐴 = 5 Since 5 < 8 < 10 , there are two solutions for the triangle.
40. 40. 40 Lesson 5.1 The Law of Sines and Its Applications SSA Case (Continuation) Example 3. Determine the number of possible solutions for each triangle. b) 𝑏 = 8, 𝑐 = 10, 𝐵 = 118 𝑜 Since 118 𝑜 ≥ 90 𝑜 , consider Case 2. In this triangle, 8 ≤ 10 , so there are no solutions.
41. 41. 41 Lesson 5.1 The Law of Sines and Its Applications SSA Case Once you have determined that there is/are one or two solution(s) for a triangle given the measures of two sides and a non-included angle, you can use the Law of Sines to solve the triangle. Example 4. Find all solutions for the triangle. If no solutions exist, write none. 𝐴 = 51 𝑜 , 𝑎 = 40, 𝑐 = 50 Since 51 𝑜 < 90 𝑜 , consider Case 1. 𝑐 𝑠𝑖𝑛 𝐴 = 50 sin 51 𝑜 𝑐 sin 𝐴 ≈ 38.85729807 Since 38.9 < 40 < 50, there are two solutions for the triangle.
42. 42. 42 Lesson 5.1 The Law of Sines and Its Applications Solution. Use the Law of Sines to find 𝐶. 𝑎 sin 𝐴 = 𝑐 sin 𝐶 40 sin 51 𝑜 = 50 sin 𝐶 40 ∙ sin 𝐶 = 50 ∙ sin 51 𝑜 sin 𝐶 = 50 sin 51 𝑜 40 𝐶 = sin−1 50 sin 51 𝑜 40 C ≈ 76.27180414 𝑜 SSA Case Example 4 (Continuation) Given: 𝐴 = 51 𝑜 , 𝑎 = 40, 𝑐 = 50
43. 43. 43 Lesson 5.1 The Law of Sines and Its Applications SSA Case Example 4 (Continuation) So, 𝐶 ≈ 76.3 𝑜 . Since we know there are two solutions, there must be another possible measurement for 𝐶. In the second case, 𝐶 must be less than 180 𝑜 and have the same sine value. Since we know that if 𝛼 < 90 𝑜, sin 𝛼 = sin 180 𝑜 − 𝛼 , 180 𝑜 − 76.3 𝑜 or 103.7 𝑜 is another possible measure for 𝐶.
44. 44. 44 Lesson 5.1 The Law of Sines and Its Applications SSA Case Example 4 (Continuation) Now solve the triangle for each possible measure of 𝐶. Solution I. 𝐵 ≈ 180 𝑜 − 51 𝑜 + 76.3 𝑜 𝐵 ≈ 52.7 𝑜 𝑎 sin 𝐴 = 𝑏 sin 𝐵 40 sin 51 𝑜 = 𝑏 sin 52.7 𝑜 b sin 51 𝑜 ≈ 40 ∙ sin 52.7 𝑜 b ≈ 40 sin 52.7 𝑜 sin 51 𝑜 𝑏 ≈ 40.94332444 One solution is 𝐵 ≈ 52.7 𝑜 , 𝐶 ≈ 76.3 𝑜 , and 𝑏 ≈ 40.9. A B C 51 𝑜 76.3 𝑜 50 b 40
45. 45. 45 Lesson 5.1 The Law of Sines and Its Applications SSA Case Example 4 (Continuation) Solution II. 𝐵 ≈ 180 𝑜 − 51 𝑜 + 103.7 𝑜 𝐵 ≈ 25.3 𝑜 𝑎 sin 𝐴 = 𝑏 sin 𝐵 40 sin 51 𝑜 = 𝑏 sin 25.3 𝑜 b sin 51 𝑜 ≈ 40 ∙ sin 25.3 𝑜 b ≈ 40 sin 25.3 𝑜 sin 51 𝑜 𝑏 ≈ 21.99627275 Another solution is 𝐵 ≈ 25.3 𝑜 , 𝐶 ≈ 103.7 𝑜 , and 𝑏 ≈ 22.0. A B C 51 𝑜 103.7 𝑜 50 b 40
46. 46. 46 Lesson 5.1 The Law of Sines and Its Applications B C A a h b c The area of any triangle can be expressed in terms of two sides of a triangle and the measure of the included angle. Suppose you know the measures of 𝐴𝐶 and 𝐴𝐵 and the measure of the included angle ∠𝐴 in ∆𝐴𝐵𝐶. Let 𝐾 represent the measure of the area of ∆𝐴𝐵𝐶, and let ℎ represent the measure of the altitude from 𝐵 . Then 𝐾 = 1 2 𝑏ℎ . But, sin 𝐴 = ℎ 𝑐 or ℎ = 𝑐 sin 𝐴 . If you substitute 𝑐 sin 𝐴 for ℎ , the result is the following formula. 𝐾 = 1 2 𝑏𝑐 sin 𝐴
47. 47. 47 Lesson 5.1 The Law of Sines and Its Applications If you drew altitudes from A and C, you could also develop two similar formulas. Area of Triangles Let ∆𝐴𝐵𝐶 be any triangle with 𝑎 , 𝑏 , and 𝑐 representing the measures of the sides opposite the angles with measurements 𝐴 , 𝐵 , and 𝐶 , respectively. Then the area 𝐾 can be determined using one of the following formulas. 𝐾 = 1 2 𝑏𝑐 sin 𝐴 𝐾 = 1 2 𝑎𝑐 sin 𝐵 𝐾 = 1 2 𝑎𝑏 sin 𝐶 B C A a h b c
48. 48. 48 Lesson 5.1 The Law of Sines and Its Applications Example 5. Find the area of ∆𝐴𝐵𝐶 if 𝑎 = 4.7 , 𝑐 = 12.4 , and 𝐵 = 47 𝑜 20′ . 𝐾 = 1 2 𝑎𝑐 sin 𝐵 𝐾 = 1 2 4.7 12.4 sin 47 𝑜 20′ 𝐾 ≈ 21.42690449 The area of ∆𝐴𝐵𝐶 is about 21.4 square units. AB C 4.7 12.4 47 𝑜20′
49. 49. 49 Deriving the Law of Cosines • Write an equation using Pythagorean theorem for shaded triangle. b h a k c - k A B C c Abk Abh cos sin         Abccba AbccAAba AbAbccAba AbcAba cos2 cos2cossin coscos2sin cossin 222 22222 222222 222    
50. 50. 50 Law of Cosines • Similarly • Note the pattern Cabbac Baccab Abccba cos2 cos2 cos2 222 222 222   
51. 51. 51 Lesson 5.2 The Law of Cosines and Its Applications Law of Cosines Let ∆𝐴𝐵𝐶 be any triangle with 𝑎, 𝑏, and 𝑐 representing the measures of the sides opposite the angles with measures 𝐴, 𝐵, and 𝐶, respectively. Then, the following are true. 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝐶 𝑏2 = 𝑎2 + 𝑐2 − 2𝑎𝑐 cos 𝐵 𝑎2 = 𝑏2 + 𝑐2 − 2𝑏𝑐 cos 𝐴 A B C b a c D a - x x h
52. 52. 52 Lesson 5.2 The Law of Cosines and Its Applications From geometry, you know that a unique triangle can be formed if a) the measures of two sides and an included angle are known (SAS) or b) the measures of three sides of a triangle are known and the sum of any two measures is greater than the remaining measure (SSS).
53. 53. 53 Lesson 5.2 The Law of Cosines and Its Applications SAS Case Example 1. Landscaping Suppose you want to fence a triangular lot as shown at the right. If two sides measure 84 feet and 78 feet and the angle between the two sides is 102 𝑜, what is the length of the fence to the nearest foot? 84 ft 78 ft 102 𝑜
54. 54. 54 Lesson 5.2 The Law of Cosines and Its Applications SAS Example 1 Solution. 𝑐2 = 𝑎2 + 𝑏2 − 2𝑎𝑏𝑐𝑜𝑠 𝐶 𝑐2 = 842 + 782 − 2 84 78 𝑐𝑜𝑠102 𝑜 𝑐2 = 7056 + 6084 + 2724.47496 𝑐2 = 15864.4748 𝑐 ≈ 125.9542568 𝑓𝑡 Let's add the three lengths. 84 𝑓𝑡 + 78 𝑓𝑡 + 125.95 𝑓𝑡 ≈ 288 𝑓𝑡 The length of the fence is about 288 ft. 84 ft 78 ft 102 𝑜
55. 55. 55 Lesson 5.2 The Law of Cosines and Its Applications SSS Case Example 2. The sides of a triangle are 18 inches, 21 inches, and 14 inches. Find the measure of the angle with the greatest measure.
56. 56. 56 Lesson 5.2 The Law of Cosines and Its Applications 18 in 21 in 14 in SSS Case (Continuation) Example 2. The sides of a triangle are 18 inches, 21 inches, and 14 inches. Find the measure of the angle with the greatest measure.
57. 57. 57 Lesson 5.2 The Law of Cosines and Its Applications Formulas in Finding for the Angles 𝐶 = cos−1 𝑎2 + 𝑏2 − 𝑐2 2𝑎𝑏 𝐵 = cos−1 𝑎2 + 𝑐2 − 𝑏2 2𝑎𝑐 𝐴 = cos−1 𝑏2 + 𝑐2 − 𝑎2 2𝑏𝑐
58. 58. 58 Lesson 5.2 The Law of Cosines and Its Applications 18 in 21 in 14 in ∠ with the greatest measure SSS Example 2 (Continuation) Solution. 𝐶 = cos−1 𝑎2 + 𝑏2 − 𝑐2 2𝑎𝑏 𝐶 = cos−1 142+182 −212 2 14 18 𝐶 = cos−1 196+324−441 504 𝐶 = cos−1 79 504 𝐶 ≈ 80.98192557 𝑜 The largest angle has a measure about 81 𝑜 .
59. 59. 59 References Holliday, B. et al. (2004) Glencoe ADVANCED Mathematical Concepts. The McGraw-Hill Companies, Inc., United States of America Senk, S. et al. (1998) UCSMP Functions, Statistics, and Trigonometry. Addison Wesley Longman, Inc., United States of America Website Link for Video http://www.youtube.com/watch?v=geDSwx2TuiE
60. 60. 60 Website Links for Images 1. https://www.google.com.ph/search?q=angles+of+elevation+and+depression&source=lnms&tbm=isc h&sa=X&ei=h -lxU9voOYOllQXK 6IHYBQ&sqi=2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555 2. https://www.google.com.ph/search?q=angles+of+elevation+and+depression&source=lnms&tbm=isc h&sa=X&ei=h- lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=solving+right+triangles&t bm=isch 3. https://www.google.com.ph/search?q=angles+of+elevation+and+depression&source=lnms&tbm=isc h&sa=X&ei=h- lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=oblique+triangles&tbm=i sch 4. https://www.google.com.ph/search?q=angles+of+elevation+and+depression&source=lnms&tbm=isc h&sa=X&ei=h- lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=acute+triangles+in+the+r eal+world&tbm=isch 5. https://www.google.com.ph/search?q=angles+of+elevation+and+depression&source=lnms&tbm=isc h&sa=X&ei=h- lxU9voOYOllQXK 6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=obtuse+triangles+in+the+ real+world&tbm=isch 6. https://www.google.com.ph/search?q=angles+of+elevation+and+depression&source=lnms&tbm=isc h&sa=X&ei=h- lxU9voOYOllQXK6IHYBQ&sqi= 2&ved=0CAYQ_AUoAQ&biw= 1252&bih=555#q=law+of+sines&tbm=isch
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Grade 9 Mathematics Module 7 Answer Key

Source: https://www.slideshare.net/DodsDodong/module-7-triangle-trigonometry-super-final

Posted by: rowestrust.blogspot.com

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