1. Right Triangles and Trigonometry
Chapter 9

2. The Pythagorean Theorem and Its Converse
I can use the Pythagorean Theorem and the Converse of the Pythagorean Theorem.

3. The Pythagorean Theorem and Its Converse
Vocabulary (page 244 in Student Journal)
Pythagorean triple: a set of nonzero whole numbers that satisfy the equation a2 + b2 = c2

4. The Pythagorean Theorem and Its Converse
Core Concepts (pages 244 and 245 in Student Journal)
Pythagorean Theorem
If a triangle is a right triangle, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.
a2 + b2 = c2

5. The Pythagorean Theorem and Its Converse
Converse of the Pythagorean Theorem
If the sum of the squares of the length of 2 sides of a triangle is equal to the square of the length of the third side, then the triangle is a right triangle

6. The Pythagorean Theorem and Its Converse
Pythagorean Inequalities Theorem
If the square of the length of the longest side of a triangle is…
greater than the sum of the squares of the lengths of the other 2 sides, then the triangle is obtuse.
less than the sum of the squares of the lengths of the other 2 sides, then the triangle is acute.

7. The Pythagorean Theorem and Its Converse
Examples (page 246 in Student Journal)
Find the value of x. Do the side lengths form a Pythagorean triple?
#2) #3)

8. The Pythagorean Theorem and Its Converse
Solutions
#2) 4√3, no
#3) 25, yes

9. The Pythagorean Theorem and Its Converse
Verify the segment lengths form a triangle. Is the triangle acute, right, or obtuse?
#8) 90, 216, 234

10. The Pythagorean Theorem and Its Converse
Solution
#8) > 234, right because = 234 2

11. Special Right Triangles
I can find side lengths in special right triangles.

12. Special Right Triangles
Core Concepts (page 249 in Student Journal)
Triangle Theorem
In a triangle, both legs are congruent and the length of the hypotenuse is the square root of 2 times the length of a leg.

13. Special Right Triangles
Triangle Theorem
In a triangle, the length of the hypotenuse is twice the length of the shorter leg and the length of the longer leg is the square root of 3 times the length of the shorter leg.

14. Special Right Triangles
Examples (page 250 in Student Journal)
Find the value of x. Write your answer in simplest form.
#1) #3)

15. Special Right Triangles
Solutions
#1) 10√2
#3) 8

16. Special Right Triangles
#6) Find the value of x and y. Write the answer in simplest form.

17. Special Right Triangles
Solution
#6) x = 11√3, y = 11

18. Similar Right Triangles
I can use geometric means.

19. Similar Right Triangles
Vocabulary (page 254 in Student Journal)
geometric mean: the positive number x that satisfies the equation a/x = x/b, where a and b are positive numbers

20. Similar Right Triangles
Core Concepts (pages 254 and 255 in Student Journal)
Right Triangle Similarity Theorem
If the altitude is drawn to the hypotenuse of a right triangle, then the 2 triangles formed are similar to the original triangle and to each other.

21. Similar Right Triangles
Geometric Mean (Altitude) Theorem
In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into 2 segments. The length of the altitude is the geometric mean of the lengths of the 2 segments of the hypotenuse.

22. Similar Right Triangles
Geometric Mean (Leg) Theorem
In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into 2 segments. The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

23. Similar Right Triangles
Examples (page 256 in Student Journal)
#1) Identify the similar triangles in the diagram.

24. Similar Right Triangles
Solution
#1) triangle IJH ~ triangle IKJ ~ triangle JKH

25. Similar Right Triangles
Find the value of the variable.
#5) #6)

26. Similar Right Triangles
Solutions
#5) x = 12
#6) y = 3√11

27. The Tangent Ratio
I can use the tangent ratio.

28. The Tangent Ratio Vocabulary (page 259 in Student Journal)
trigonometric ratio: equivalent ratios created from the corresponding sides of similar right triangles
tangent (tan): the ratio of the length of the opposite leg to the length of the adjacent leg

29. The Tangent Ratio
angle of elevation: the angle above a horizontal line

30. The Tangent Ratio Examples (pages 260 and 261 in Student Journal)
#1) Find the tangents of the acute angles.

31. The Tangent Ratio
Solution
#1) tan <R = 15/8, tan <S = 8/15

32. The Tangent Ratio
#4) Find the value of x.

33. The Tangent Ratio
Solution
#4) x = 28.4

34. The Tangent Ratio
#9) A boy flies a kite at an angle of elevation of 18 degrees. The kite reaches its maximum height 300 feet away from the boy. What is the maximum height of the kite?

35. The Tangent Ratio
Solution
#9) 97.5 feet

36. The Sine and Cosine Ratios
I can use the sine and cosine ratios.

37. The Sine and Cosine Ratios
Vocabulary (page 264 in Student Journal)
sine (sin): the ratio of the length of the opposite leg to the length of the hypotenuse
cosine (cos): the ratio of the length of the adjacent leg to the length of the hypotenuse

38. The Sine and Cosine Ratios
angle of depression: the angle below a horizontal line

39. The Sine and Cosine Ratios
Core Concepts (pages 264 and 265 in Student Journal)
SOH-CAH-TOA
Sine and Cosine of Complementary Angles
The sine of an acute angle is equal to the cosine of its complement and vice versa.

40. The Sine and Cosine Ratios
Examples (pages 265 and 266 in Student Journal)
#1) Find sin F, sin G, cos F, cos G.

41. The Sine and Cosine Ratios
Solution
#1) sin F = 12/13, sin G = 5/13
cos F = 5/13, cos G = 12/13

42. The Sine and Cosine Ratios
Find the value of each variable.
#10)

43. The Sine and Cosine Ratios
Solution
#10) x = 2.5, y = 8.7

44. Solving Right Triangles
I can use inverse trigonometric ratios to solve right triangles.

45. Solving Right Triangles
Vocabulary (page 269 in Student Journal)
inverse tangent (tan-1): if tan A = x, then
tan-1x = measure of angle A
inverse sine (sin-1): if sin A = y, then
sin-1y = measure of angle A

46. Solving Right Triangles
inverse cosine (cos-1): if cos A = z, then
cos-1z = measure of angle A
solve a right triangle: finding all unknown side lengths and angle measures in a right triangle

47. Solving Right Triangles
Examples (pages 270 and 271 in Student Journal)
#7) Solve the right triangle.

48. Solving Right Triangles
Solution
#7) AC = 13.4, m<A = 63.4 degrees,
m<C = 26.6 degrees

49. Solving Right Triangles
#11) A boat is pulled by a winch on a dock 12 feet above the dock of the boat. When the winch is fully extended to 25 feet, what is the angle of elevation from the boat to the winch?

50. Solving Right Triangles
Solution
#11) 28.7 degrees

51. Law of Sines and Law of Cosines
I can use the Law of Sines and the Law of Cosines to solve triangles.

52. Law of Sines and Law of Cosines
Vocabulary (page 274 in Student Journal)
Law of Sines: used to solve triangles when 2 angles and the length of any side are known (AAS or ASA), or when the lengths of 2 sides and an angle opposite one of those 2 sides are known (SSA)

53. Law of Sines and Law of Cosines
Law of Cosines: used to solve triangles when 2 sides and the included angle are known (SAS), or when all 3 sides are known (SSS)

54. Law of Sines and Law of Cosines
Core Concepts (pages 274 and 275 in Student Journal)
Area of a Triangle
The area of any triangle can be found by ½ the product of the lengths of any 2 sides times the sine of their included angle.

55. Law of Sines and Law of Cosines
Law of Sines Theorem
sin A/a = sin B/b = sin C/c, where a, b and c are side lengths opposite angles A, B and C respectively

56. Law of Sines and Law of Cosines
Law of Cosines Theorem
If a, b and c are side lengths of a triangle opposite angles A, B and C respectively, then the following are true:
a2 = b2 + c2 – 2bc(cos A)
b2 = a2 + c2 – 2ac(cos B)
c2 = a2 + b2 – 2ab(cos C)

57. Law of Sines and Law of Cosines
Examples (page 276 in Student Journal)
#4) Find the area of the triangle.

58. Law of Sines and Law of Cosines
Solution
#4) 61.8 units squared

59. Law of Sines and Law of Cosines
#6) Solve the triangle.

60. Law of Sines and Law of Cosines
Solution
#6) m<C = 100 degrees, AB = 43.3, BC = 33.7