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Chapter 9

Chapter 9. Right Triangles and Trigonometry. Chapter 9 Objectives. Prove Pythagorean Theorem Utilize Pythagorean converse Identify right, obtuse, and acute triangles using Pythagorean converse Define properties of a 45-45-90 triangle Define properties of a 30-60-90 triangle

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Chapter 9

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  1. Chapter 9 Right Triangles and Trigonometry

  2. Chapter 9 Objectives • Prove Pythagorean Theorem • Utilize Pythagorean converse • Identify right, obtuse, and acute triangles using Pythagorean converse • Define properties of a 45-45-90 triangle • Define properties of a 30-60-90 triangle • Apply trigonometry to geometric situations • Utilize inverse trigonometry functions

  3. Lesson 9.2 The Pythagorean Theorem

  4. a b Theorem 9.4:Pythagorean Theorem • In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. • c2 = a2 + b2 • cis always the hypotenuse • a and b are the legs in any order c

  5. b a a c b c c b c a a b -2ab -2ab Proof of Pythagorean Theorem How would you find the area of the figure at right? (Hint: 2 ways) Side length squared Area of triangles plus area of inner square (a + b)2 Should give same answer 4(1/2ab) + c2 = (a + b)(a + b) 2ab + c2 a2 + 2ab + b2 a2 + b2 = c2

  6. Example 1 Find the length of the missing side • a = 3, b = 3 • c2 = a2 + b2 c2 = 32 + 32 c2 = 9 + 9 c2 = 18 c = 18 c = 32  4.2 • a = 5, c = 13 • c2 = a2 + b2 132 = 52 + b2 169 = 25 + b2 b2 = 169 - 25 b2 = 144 b = 144 b = 12 c2 = a2 + b2

  7. Pythagorean Triple • A Pythagorean triple is a set of three positive integers that satisfy the Pythagorean Theorem. • That means that a,b, and c will all be nice whole numbers • except 0 • 3,4,5 • 5,12,13

  8. Example 2 Find the length of the missing side and determine if the sides form a Pythagorean Triple • a = 18, b = 24 • c2 = a2 + b2 c2 = 182 + 242 c2 = 324 + 576 c2 = 900 c = 900 c = 30 Yes, a Pythagorean Triple is formed! c2 = a2 + b2

  9. Lesson 9.3 Converse of the Pythagorean Theorem

  10. 3 4 4 3 2 1 6 6 6 Theorem 5.13:Triangle Inequality • The sum of the lengths of any two sides of a triangle is greater than the length of the third side. Add each combination of two sides to make sure that they are longer than the third remaining side.

  11. Theorem 9.5:Converse of thePythagorean Theorem • If c2 = a2 + b2 is true, then the triangle in question is a right triangle. • You need to verify the three sides of the triangle given will make the Pythagorean Theorem true when plugged in. • Remember the largest number given is always the hypotenuse • Which is c in the Pythagorean Theorem

  12. Example 3 Determine whether the following triangles are right triangles or not. • 4,7,10 • c2 = a2 + b2102 = 42 + 72100 = 16 + 49100  65NO • 5,12,13 • c2 = a2 + b2132 = 52 + 122169 = 25 + 144169 = 169YES

  13. c a b Theorem 9.6:Acute Triangles fromPythagorean Theorem • If c2 < a2 + b2, then the triangle is an acute triangle. • So when you check if it is a right triangle and the answer for c2 is smaller than the answer for a2 + b2, then the triangle must be acute • It essentially means the hypotenuse shrunk a little! • And the only way to make it shrink is to make the right angle shrink as well!

  14. c a b Theorem 9.7:Obtuse Triangles fromPythagorean Theorem • If c2 > a2 + b2, then the triangle is an obtuse triangle. • So when you check if it is a right triangle and the answer for c2 is larger than the answer for a2 + b2, then the triangle must be obtuse • It essentially means the hypotenuse grew a little! • And the only way to make it grow is to make the right angle grow as well!

  15. Example 4 Determine if the following sides create a right, obtuse, acute, or no triangle at all. A) 38, 77, 86 B) 10.5, 36.5, 37.5 c longest side Triangle Y/N Triangle Y/N c2 = a2 + b2 c2 = a2 + b2 862 = 382 + 772 37.52 = 10.52 + 36.52 7396 = 1444 + 5929 1406.25 = 110.25 + 1332.25 7396 = 7373 1406.25 = 1442.5 7396 > 7373 1406.25 < 1442.5 acute obtuse

  16. Lesson 9.5 Trigonometric Ratios

  17. Trigonometric Ratios • A trigonometric ratio is a ratio of the lengths of any two sides in a right triangle. • You must know: • one angle in the triangle other than the right angle • one side (any side) of the triangle. • These help find any other side of the triangle.

  18. c a b = Sine • The sine is a ratio of • side opposite the known angle, and… • the hypotenuse • Abbreviated • sin • This is used to find one of those sides. • Use your known angle as a reference point θ b side opposite θ sin θ = c hypotenuse

  19. c a b = Cosine • The cosine is a ratio of • side adjacent the known angle, and… • the hypotenuse • Abbreviated • cos • This is used to find one of those sides. • Use your known angle as a reference point θ a side adjacent θ cos θ = c hypotenuse

  20. c a b = Tangent • The tangent is a ratio of • side opposite the known angle, and… • side adjacent the known angle • Abbreviated • tan • This is used to find one of those sides. • Use your known angle as a reference point θ b side opposite θ tan θ = side adjacent θ a

  21. SOHCAHTOA S o h C a h T o a in pposite • This is a handy way of remembering which ratio involves which components. • Remember to start at the known angle as the reference point. • Also, each combination is a ratio • So the sin is the opposite side divided by the hypotenuse ypotenuse os djacent ypotenuse an pposite djacent

  22. 7 x 37o x 4 42o sin 42o = cos 37o = 4 = .7986 Example 5 • First determine which trig function you want to use by identifying the known parts and the variable side. • Use that function on your calculator to find the decimal equivalent for the angle. • Set that number equal to the ratio of side lengths and solve for the variable side using algebra. If you do not have a calculator with trig buttons, then turn to p845 in book for a table of all trig ratios up to 90o. Get x out of denominator first by multiplying both sides by x. x 4 7 x 7 (sin 42o) = x x (cos 37o) = 4 4 7 (.6691) = x = 4.683 x = = 5.008 cos 37o

  23. Lesson 9.4 Special Right Triangles

  24. 45o 1 45o 1 Theorem 9.8:45-45-90 Triangle • In a 45-45-90 triangle, the hypotenuse is 2 times as long as each leg. • Remember each leg is the same length • Theorem 4.7: Converse of the Base Angles  2

  25. If you are needing to find the length of the hypotenuse, simply multiply the length of either leg by 2. Best way to do that is simply write the leg followed by 2. When you know the hypotenuse, find the number in front of the 2 and that is the length of your leg. If there is no 2, then you need to divide by 2 on your calculator. 45o 45o  2 6 6 4 45o 45o 6 4 Example 6 I II  2 4 Find the length of the missing side(s)

  26. 60o 30o Theorem 9.9:30-60-90 Triangle • In a 30-60-90 triangle • the hypotenuse is twice the length of the shortest leg, and… • the longer leg is the 3 times as long as the shortest leg. 2 1  3

  27. Knowing the shortest leg and trying to find the hypotenuse, simply multiply the shortest leg by 2. If you need to find the shortest leg, then divide the hypotenuse by 2. Knowing the shortest leg and trying to find the longer leg, multiply the shortest leg by 3. Or just write 3 after the length of the shortest leg. 60o 30o Example 7 8 4 4  3 Find the length of the missing side(s)

  28. Lesson 9.6 Solving Right Triangles

  29. Inverse Trig Ratios Inverse trig ratios are used to find the measure of the angles of a triangle. The catch is…you must know two side lengths. Those sides determine which ratio to use based on the same ratios we had from before. SOHCAHTOA

  30. θ 7 17 4 9 4 4 9 9 θ 7 7 17 17 Example 8 • You still base your ratio on what sides are you working with compared to the angle you want to find. • Only now, your variable is θ. • So once you find your ratio, you will then use the inverse function of your ratio from your calculator sin θ = cos θ = θ = sin-1 θ = cos-1 θ = sin-1 .5714 θ = cos-1 .5294 θ = 34.8o SOHCAHTOA θ = 58.0o

  31. Solving a Triangle • To solve a right triangle, you must find • all 3 sides • all 3 angles • or the other 2 angles besides the right angle • So your final answer when solving a right triangle will have six parts to the answer!

  32. Lesson 9.7 Vectors

  33. Vector Magnitude • The magnitude of a vector, AB, is the distance from the initial pointA to the terminal point B. • It is written as AB • Because the magnitude of a vector is essentially finding the distance between two points, we can use the Distance Formula to calculate the magnitude of the vector. • AB = [(x2 – x1)2 + (y2 – y1)2] • Or if you know the component form of the vector, you can use the Pythagorean Theorem to calculate the magnitude. • x2 + y2 = AB2

  34. Example 9 Calculate the magnitude of the vector AB. • A(0,0) , B(4,5) • AB = [(x2 – x1)2 + (y2 – y1)2] AB = [(4 – 0)2 + (5 – 0)2] AB = [42 + 52] AB = [(16 + 25] AB = 41  6.4 • AB = <-4,-2> • x2 + y2 = AB2 (-4)2 + (-2)2 =AB2 16 + 4 =AB2 AB2 = 20 AB = 20  4.5

  35. Direction of a Vector • The direction of a vector, , is determined by the angle it makes with a horizontal line that intersects the initial pointA. • Since the vector has a horizontal and vertical component, we can fill in the triangle to use a trigonometric ratio in calculating the direction of a vector. • Knowing the components of the vector leave us with the inverse tangent to calculate the necessary angle •  = tan-1y/x • Two vectors are equal if they have the same magnitude and direction. • However, they need not share the same initial or terminal points. • Two vectors are parallel if they have the same or opposite direction. • The opposite direction would be 180o different than the original.

  36. Adding Vectors • The sum of any two vectors is the individual sums of the horizontal and vertical components. u = <a,b> v = <c,d> u+ v = <a+c,b+d>

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