Presentation is loading. Please wait.

Presentation is loading. Please wait.

Special Right Triangles

Similar presentations


Presentation on theme: "Special Right Triangles"— Presentation transcript:

1 Special Right Triangles
Unit 4

2 Simplifying Radicals √ radical Radicand – number inside the radical
You can click on other videos for more explainations.

3 Examples √6 ∙ √8 √2∙2∙2∙2∙3 4√3 2) √90 √2∙3∙3∙5 3√10
√6 ∙ √8 √2∙2∙2∙2∙3 4√3 2) √90 √2∙3∙3∙5 3√10 3) √243 √3 √3∙3∙3∙3∙3 9 √3 9

4 Division – multiply numerator and denominator by the radical in the denominator
4) √25 √3 5 ∙√3 √3 ∙√3 3 = √14 6) √5 ∙ √35 √14 √5∙5 ∙7 √2∙7 5√7 √2∙7 √2∙7 √2∙7 35 √2 = 5 √2 14 2

5 What have you learned today? What is still confusing?

6 Pythagorean Theorem and It’s Converse
Objective: to use the Pythagorean Theorem and it’s converse. c2 = a2 + b2

7 Key Concept Pythagorean Theorem c2 = a2 + b2 c - hypotenuse
a – altitude leg b – base leg

8 Key Concepts Acute c2 < a2 + b2 Right c2 = a2 + b2
C b A Acute c2 < a2 + b2 Right c2 = a2 + b2 Obtuse c2 > a2 + b2 B a c C b A B a c C b A

9 Ex 1 Find the value of x. Leave in simplest radical form.
Answer: 2 √11 x 12 10

10 Ex 2: Baseball A baseball diamond is a square with 90 ft sides. Home plate and second base are at opposite vertices of the square. About far is home plate from second base to the nearest foot? About 127 ft

11 Ex 3:Classify the triangle as acute, right or obtuse.
15, 20, 25 right b) 10, 15, 20 Obtuse

12 Pythagorean Triplet Whole numbers that satisfy c2 = a2 + b2. Example: 3, 4, 5 Can you find another set?

13 What have you learned today? What is still confusing?

14 Special Right Triangles
Objective: To use the properties of 45⁰ – 45⁰ – 90⁰ and 30⁰ – 60⁰ - 90⁰ triangles.

15 Isosceles Right Triangle Key Concept
45⁰-45⁰ -90⁰ X – X – X2 102 22 2 10 2 10

16 Example 1 Find x. Simplify. 45 – 45 – 90 x – x - x 2 8 – 8 – 82 82

17 Try 1: FIND THE MISSING LENGTHS. SIMPLEST RADICAL FORM.
45⁰ 45⁰ 90⁰ X X X2 1212 X2 = 1212 2 2 X = 121 MISSING SIDE LENGTHS ARE 121. 1212 45⁰

18 Example 2 Find x. Simplify. 45 – 45 – 90 y – y - y 2 – – = y 2 set up equation 2 2 divide by root 2 y = 28 2 2 = 28 2 2 = 142

19 Try2: Find the length of the hypotenuse of a 45⁰ – 45⁰ – 90⁰ triangle with legs of length 5√6 . 45⁰ – 45⁰ – 90⁰ x - x - x√2 X = 5√6 x√2 = 5√6√2 substitute into the formula = 10 √3

20 Try 3 Find the length of a leg of a 45⁰ – 45⁰ – 90⁰ triangle with hypotenuse of length ⁰ – 45⁰ – 90⁰ x - x - x√2 x√2 = 22 solve for x X = 22 = 22√2 = 11√2 √2 2

21 Try 4: The distance from one corner to the opposite corner of a square field is 96ft. To the nearest foot, how long is each side of the field? 45⁰ – 45⁰ – 90⁰ x - x - x√2 x√2 = 96 solve for x X = 96 = 96√2 = 48√2 √2 2

22 What is the relationship of the legs and hypotenuse of an isosceles right triangle?
45⁰ – 45⁰ – 90⁰ x - x - x√2

23 Equilateral Triangle Key Concept
30⁰ - 60⁰ - 90⁰ X – X3 – 2X 30⁰ 4 2 3 60⁰ 2

24 Example 3 Solve for missing parts of each triangle: x = 10 y = 5√3 x y

25 Example 4 The longer leg of a 30⁰ – 60⁰ - 90⁰ triangle has length of 18. Find the lengths of the shorter leg and the hypotenuse. 30⁰ – 60⁰ - 90⁰ x - x√3 - 2x x√3 = 18 solve for x X = 18 = 18√3 = 6√3 – short leg √3 3 12√3 - hypotenuse

26 Try 5 X - X3 - 2X 9 2X = 9 X = 9/2 a = 9/2 b = 93 2 X - X3 - 2X = x 3 3 3 X = 25 3 3 d = 25 3 c = 50 3 3

27 What is the relationship of the 30-60-90 right triangle?
30⁰ – 60⁰ - 90⁰ x - x√ x

28 Exit Ticket X = 5 Y = 5 2 Z = 53 3 W = 103

29 What have you learned today? What is still confusing? Click on the link.

30 Similarities in Right Triangles
Objective: To find and use relationships in similar right triangles

31 Geometric mean with similar right triangles

32 Type 1 Relationship: altitude side2 side 1 altitude = altitude Side 1

33 Example 1

34 Try 1:

35 Type 2 relationship: Hypotenuse Leg(hyp) Leg ______ side = Big Small

36 Example 2:

37 Example 3:

38 Try: 2

39 Try 3:

40 Try 4:

41 Exit Ticket Find x, y, and z. X = 6 9 x 36 = 9x 4 = x 9 = z z 9+x
y = x 9+x y Y ² = 4(13) Y = 2√13

42 8-3 and 8-4 Right Triangle Trigonometry
Objective To use sine, cosine and tangent ratios to determine side and angle measures in triangles

43 Key Concept Sine (sin) Cosine(cos) Tangent(tan) SOH CAH TOA The Old Aunt Sat On Her Coat and Hat

44 Example 1 Find sin E, cos E, and tan E. sin E = 8/10 = 4/5
Try: Find sin F, cos F, and tan F. sin F = 6/10 = 3/5 cos F = 8/10 = 4/5 tan F =6/8 = 3/4 E G F

45 Example 2 – round to the nearest tenths.
OPP/ADJ Tan (37⁰) = x/3 3(tan (37⁰)) = x X 2.3 Opp adj

46 Try 1 OPP/HYP Sin (67⁰) = 10/x 10/sin (67⁰) = x X  10.9 Opp hyp

47 TRY 2 Adj/hyp Cos (40⁰) = 6/x 6/cos(40⁰) = x X 4.6 adj hyp

48 Ex 3:When solving for angles use inverse. Round to the nearest degree
Adj/hyp Cos(x) = 8/18 Cos-1(8/18)  64⁰

49 TRY Opp/adj Tan (x) = 11/8 tan-1(11/8)  54⁰

50 What have you learned today? What is still confusing?

51 Angles of Elevation and Depression
Objective: To use angles of elevation and depression to solve problems

52 Try Round to the nearest tenths.
Opp/adj Tan(25⁰) = x/ (tan(25⁰)) = x x 116.6 ft

53 Try round to the nearest hundredth
Opp/hyp Sin (40⁰) = 30/x 30/sin(40⁰) = x X 46.67 ft

54 Try Opp/adj = tan α Tan(35⁰) = x / 30 30(tan(35⁰)) = x X 52
= 57ft X 5ft 35⁰ 30 ft

55 What have you learned today? What is still confusing? Click on the link.

56 Law of Sines Objective: To derive the Law of Sines
To apply the law of sines to solve problems

57 Law of Sines Sin A Sin B Sin C a b c When to use:
Oblique triangles given: AAS - 2 angles and 1 adjacent side ASA - 2 angles and their included side = =

58 Example 1: Find the missing angle and sides.
The angles in a ∆ total 180°, so angle C = 30°. Set up the Law of Sines to find side b: A C B 70° 80° a = 12 c b

59 Continued A C B 70° 80° a = 12 c b = 12.6 30°

60 Try 1 Angle A = 35° Side b = 26.2 cm Side c = 47.4 cm 30° A C B 115°

61 When do you use Law of Sines? What is still confusing?

62 Law of Cosines Objective: To derive the Law of Cosines
To apply the law of cosines to solve problems

63 A (bcos C, bsinC) b c (0, 0) B a C (a, 0)

64 A b c A (bcos C, bsinC) B a C b c (0, 0) B a C (a, 0) (bcos C, bsinC)

65 (a, 0) (bcos C, bsinC) C A B c b a

66 Law of Cosines

67 Law of Cosines When you use it:
SAS – when you know two sides and an included angle and you want to find the third side SSS – when you know three sides and you want to find an angle

68 Example 1 A Find the missing sides and angles 7 c 30 C B 4

69 Try 1: A Find the missing sides and angles. c 21 123º B C 18

70 C The leading edge of each wing of the B-2 Stealth Bomber measures feet in length. The angle between the wing's leading edges is °. What is the wing span (the distance from A to C)? A

71 Example 2: Find the missing angles.
C B a = 30 c = 15 b = 20

72 Try 2:

73 When do you use Law of Cosines? What is still confusing?

74 Objective: To use and apply vectors to solve applied problems.

75 How to draw and read a vector.
V U Tail Head O OA A

76 How to Draw the Sum of Vectors
  U + V V U O U OA A

77 Try: Draw the sum of and U V V U

78 How to sketch a vector on a coordinate plane

79 Try: Sketch the vectors below.

80 How to find the magnitude of a vector (Simplest radical form)

81 How to find the direction of a vector
To find the angle we will need to use Trigonometric ratios. tan (α) = 15/10 Tan-1(15/10)  56⁰ 15√ 10

82 Try: Resultant (magnitude and direction)

83 How do you find the direction and magnitude of a vector
How do you find the direction and magnitude of a vector? What is still confusing?


Download ppt "Special Right Triangles"

Similar presentations


Ads by Google