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Section 8.1

I can use the Pythagorean Theorem to find lengths of sides of triangles. I can use the Pythagorean Theorem to determine if a triangle is a right triangle. Section 8.1. hypotenuse. leg. leg. c. b. a 2 + b 2 = c 2. a. a 2 + b 2 = c 2. a 2 + b 2 = c 2. 20 2 + 30 2 = x 2.

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Section 8.1

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  1. I can use the Pythagorean Theorem to find lengths of sides of triangles I can use the Pythagorean Theorem to determine if a triangle is a right triangle Section 8.1

  2. hypotenuse leg leg c b a2 + b2 = c2 a

  3. a2 + b2 = c2 a2 + b2 = c2 202 + 302 = x2 x2 + 42 = 92 400 + 900 = x2 x2 + 42 = 92 1300 = x2 x2 + 16 = 81 x2 = 65

  4. right triangle whole numbers

  5. a2 + b2 = c2 a2 + b2 = c2 82 + 62 = 102 72 + 92 = 112   64 + 36 = 100 49 + 81 = 121 Yes, right triangle No, not a right triangle Yes, Pythagorean triple No, not a Pythagorean triple

  6. a2 + b2 = c2 skip  24 + 25 = 49 Yes, right triangle No, not a Pythagorean triple

  7. a2 + b2 = c2  8 + 32 = 40 Yes, right triangle

  8. Section 8.2 – 45-45-90Triangles • I can use properties of 45-45-90 triangles to label missing sides.

  9. 45 Given the two sides of the right triangle, use Pythagorean Theorem to find the missing side… x x2 + x2 = c2 45 2x2 = c2 x NOTE: These ratios will ALWAYS work for 45-45-90 triangles!

  10. Find the missing sides, given x: x 7 x 4 7 x 4 x = 4, so what are the other two sides? x x = 7, so what are the other two sides?

  11. Find the missing side, given the hypotenuse: 6 x x = 6, so what are the other two sides? 6 x

  12. Your Turn… x x

  13. Your Turn y y

  14. 5 If the perimeter is 20, how big is each side? 45 5 5 5 What kind of “special” triangle is formed by the picture? 45 x = 5, so how big is the diagonal? 5 x x

  15. Your Turn… Example 5! 12 x x

  16. Warm up • Find the length of the diagonal of a square with a perimeter of 20 cm. 2. Find the perimeter of a square with a diagonal of length 12 cm.

  17. Section 8.3 – 30-60-90 Triangles • I can use properties of 30-60-90 triangles to label missing sides.

  18. Your Turn

  19. Warm Up

  20. Warm Up Answers

  21. Section 8.4 - Trigonometry Objectives: • Find trigonometric ratios using right triangles • Solve problems using trigonometric ratios

  22. Ratio • A ratio is a comparison of two amounts. • Example: There are 12 boys and 11 girls in this class. What is the ratio of boys to girls? or 12:11

  23. Trigonometry • Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles.

  24. Trigonometric Ratio • Atrigonometric ratio is a ratio of the lengths of the sides of a right triangle.The 3 most common trigonometric ratios are sine, cosine, and tangent.

  25. B A C SOH CAH TOA • sine: • cosine: • tangent: sin Ao = cos Ao = tan Ao =

  26. SOH CAH TOA

  27. F 17 8 D E 15 Example 1 • Find the following trig ratios. Write your answer as a reduced fraction. • sin D • cos D • tan D

  28. F 20 12 D E 16 Example 2 • Find the following trig ratios. Write your answer as a reduced fraction. • sin F • cos F • tan F

  29. 26 Your Turn • Find the following trig ratios. Write your answer as a reduced fraction. sin J cos J tan J H 24 G J 10

  30. Example 3 – CALCULATOR IN DEGREE MODE!!! Use a calculator to find the following values to the nearest hundredth (2 decimal places). • a. sin 47o b. cos 32o c. tan 84o

  31. Example 4 • Find the value of x. Round to the nearest hundredth. x

  32. x x

  33. Your Turn x x

  34. Section 8.5 – TrigonometrySolving for an Angle Objectives: • Find trigonometric ratios using right triangles • Solve problems using trigonometric ratios

  35. Example 1 • Use a calculator to find the measure of each angle to the nearest degree. a. sin K = 0.5150 b. tan M = 7.1154 c. cos R = 0.2756

  36. Example 2 • Find the missing angle measure in each triangle to the nearest degree.

  37. Example 3

  38. I can solve problems involving angle of elevation and depression using SOH CAH TOA Section 8.6 – Angles of Elevation and Depression

  39. intersecting line of sight intersecting

  40. H x 78° 5 A 5 cos 78ᵒ = x 1 5 = x • cos 78 x = 24 feet

  41. O 12.5 x° 18 A 12.5 tan x = 18 x = tan-1(12.5/18) = 34.7ᵒ

  42. O 40 13.25° x A 40 tan 13.25ᵒ = x 1 40 = x • tan 13.25 x = 170 feet

  43. 1000 208 x°

  44. 127 132 ? x° 100 5

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