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7.1 & 7.2 Law of Sines

C. C. a. b. a. b. B. B. A. c. c. A. 7.1 & 7.2 Law of Sines. Oblique triangle – A triangle that does not contain a right angle. sin A sin B sin C a b c. =. =. or. a b c__ sin A sin B sin C. =. =. C. b. a. B. A. 14cm.

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7.1 & 7.2 Law of Sines

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  1. C C a b a b B B A c c A 7.1 & 7.2 Law of Sines Oblique triangle – A triangle that does not contain a right angle. sin Asin Bsin C a b c = = or a b c__ sin A sin B sin C = =

  2. C b a B A 14cm Solving SAA or ASA Triangles SAA and ASA Triangles – Oblique triangles in which you know 2 angles and 1 side. sin Asin Bsin C a b c a b c__ sin A sin B sin C = or = = = Solving for <B There are 180 degrees in a triangle, so 64 + 82 + B = 180 146 + B = 180 B = 34° 82° Solving for a Solving for b Sin (82) = Sin (64)Sin(82) = sin(34) 14 a 14 b aSin(82) = 14 Sin(64) a = 14 Sin(64) Sin (82) a = 12.7cm b = 7.9 cm 64°

  3. C a 20 B c A Try this one on your own sin Asin Bsin C a b c a b c__ sin A sin B sin C = = or = = 102° 38° Check your answers: < C = 40° a = 31.8 c = 20.9

  4. SSA Ambiguous Case Triangles SSA Triangles – Oblique triangles in which you know 2 of the sides and an angle opposite one of the known sides. You may have NO solutions, ONE solution, or TWO solutions sin Asin Bsin C a b c Example (NO solutions) Solve the triangle ABC if A = 75°, a = 51, and b = 71

  5. SSA Ambiguous Case(1 & 2 Solution Triangles) sin Asin Bsin C a b c Example (TWO solutions) Solve the triangle ABC A = 40°, a = 54, and b = 62 Example (ONE solution) Solve the triangle ABC A = 61°, a = 55, and c = 35

  6. C b a B A c Area of Oblique Triangles sin A = h b h = b SinA Since Area = ½ (Base)(Height) Area = ½ c (b SinA) Area = ½ b c Sin A h You may use any one of 3 area formulas. Area = ½ b c Sin A Area = ½ a b Sin C Area = ½ a c Sin B Example: (Find the Area) A = 62° b = 10 c = 24 m

  7. C C a b a b B B A c c A 7.3 Law of Cosines a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C

  8. C b=20 a B A c=30 Solving SAS Triangles Law of Cosines a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C 60° Law of Sines sin Asin Bsin C a b c Step 1: Use Law of Cosines to find side opposite given angle a2 = b2 + c2 – 2bc cos A a2 = (20)2 + (30)2 – 2(20)(30) cos (60) a = 26 Step 2: Use Law of Sines to find angle opposite the shorter of the 2 given sides. Step 3: Find the 3rd angle in the triangle using concept of 180 degrees in a triangle. = = sin B = sin (60) 20 26 B = 41° C = 79°

  9. C a 7 B 8 A Try this one on your own Law of Cosines a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C 120° Law of Sines sin Asin Bsin C a b c Check your answers: < B = 28° a = 13 <C = 32°

  10. C b=9 a=6 A c=4 B Solving SSS Triangles Law of Cosines a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C Step 1: Use Law of Cosines to find angle opposite the longest side b2 = a2 + c2 – 2ac cos B 2ac cos B = a2 + c2 - b2 cos B = a2 + c2 - b2 2ac cos B = 62 + 42 – 92 = -29/48 B = 127.2 2(6)(4) Step 2: Use Law of Sines to find either of the other angles. A = 32.1 Step 3: Find the 3rd angle in the triangle using concept of 180 degrees in a triangle. C = 20.7 Law of Sines sin Asin Bsin C a b c = =

  11. C b=16 a=10 A c=8 B Try this one on your own Law of Cosines a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C Law of Sines sin Asin Bsin C a b c = =

  12. A c=2427 mi b=331 miles C B a=2451 miles Heron’s Formula for Triangle Area(Derived from Law of Cosines) If a triangle has sides of lengths a, b, c with semiperimeter s = (1/2)(a + b + c) Area = s(s - a)(s - b)(s – c) Example:Find the area of the triangular region: S = (1/2)(2451+331+2427) = 2604.5 A = 2604.5 (2604.5 –2451)(2604.5 –331)(2604.5 –2427) A = 401,700 mi2

  13. 7-4 & 7.5 Vectors • Scalar – A quantity that involves magnitude, but no direction. • (Example: the temperature outside might be 75°) • Vector– A quantity that involves both magnitude and direction. • (Example: The jet is flying 120 mph 30° East of North) • Magnitude – How big something is – the size • Direction – The angle in degrees something is travelling • Vectors can be represented as directed line segments • Magnitude = length of segment (distance formula) • ||v|| or | v| - the magnitude of vector v v u

  14. Showing 2 Vectors are Equal • Two vectors (u and v) are equal if they have the same magnitude • and the same direction. • Magnitude • Use the distance formula to see if ||u|| = ||v|| • Direction • Find the slope of each vector. If the slopes are the same • u and v have the same direction. • Example • Let u have initial point (-3, -3) and terminal point (0, 3) • Let v have initial point (0, 0) and terminal point (3, 6) • Does u = v? YES! - ||v|| = ||u|| = 45 = 35 and both slopes = 2

  15. Scalar Multiplication (Geometric) v 2v ½ v -2v Multiplying a vector by a positive number changes the magnitude but not the direction Multiplying a vector by a negative number reverses the direction If k is a real number and v is a vector, then kv is a scalar multiple of vector v.

  16. Resultant vector u + v Terminal point of v v u Initial point of u The Geometric Method for Adding Two Vectors A geometric method for adding two vectors is shown below. The sum of u + v is called the resultant vector. Here is how we find this vector. 1. Position u and v so the terminal point of u extends from the initial point of v. 2. The resultant vector, u + v, extends from the initial point of u to the terminal point of v. Opposite vectors: v and –v -- Same magnitude, opposite direction -- The sum of opposite vectors has magnitude 0 – called the zero vector v -v

  17. y y 1 1 j j x x i i O O 1 1 Vectors in the X/Y Axes Position Vector: Vector with initial point at origin Position Vector, v, with endpoint at (a,b) is written <a, b> i and j are unit vectors (they have magnitude of 1) i = <0, 1> and j = <1, 0> • Vector i is the unit vector whose direction is along the positive x-axis. Vector j is the unit vector whose direction is along the positive y-axis. Vectors are also represented in terms of i and j. A vector, v, with initial point (0,0) and termianl point (a,b) is also represented <a, b> Magnitude: v=ai+bj (a, b) bj  a a = ||u|| cos  b = ||u|| sin  v = <a, b> = <||u|| cos , ||u|| sin > (Pythagorean Theorem)

  18. Terminal point 5 4 3 v = -3i + 4j 2 1 -5 -4 -3 -2 -1 1 2 3 4 5 -1 -2 -3 Initial point -4 -5 General formula for representation of a vector v with initial point P1 = (x1, y1) and terminal point P2 = (x2, y2) is equal to the position vector <(x2 – x1) , (y2 – y1) > and v = (x2 – x1)i + (y2 – y1)j. Vector Representation & Examples Example 1: Sketch the vector <-3. 4> , v = -3i + 4j and find its magnitude. Example 2: Vector v has initial point (3, -1) and terminal point (-2, 5) • Represent it in i, j format b) Represent it in <x , y> form c) Find magnitude ||v|| v = (-2 – 3)i + (5 – (-1))j v = -5i + 6j <-5, 6> 61

  19. Vector Operations: Addition, Subtraction, and Scalar Multiplication v = a1i + b1j and w = a2i + b2j v + w = (a1 + a2)i + (b1 + b2)j v – w = (a1 – a2)i + (b1 – b2)j Example: v = 5i + 4j and w = 6i – 9j v + w = __________________ v – w = ___________________ -i +13j 11i – 5j Let k be a constant and v = a1i + b1j then kv = (ka)i + (kb)j Example: v = 5i + 4j 6v = ___________________ -3v = __________________ -15i - 12j 30i + 24j

  20. Vector Operations Revisited v = <a,b> and w = <c, d> v + w = <(a + c), (b + d)> v – w = <(a – c), (b – d)> Example: v = <5, 4> and w = <6, -9> v + w = __________________ v – w = ___________________ <11 – 5> <-1, 13> Let k be a constant and v =< a , b> then kv = <(ka), (kb)> Example: v = <5, 4> 6v = ___________________ -3v = __________________ <30, 24> <-15, – 12> If v = <a, b> then –v = <-a, -b>  Example: v = <6, -2>, so –v = <-6, 2>

  21. Zero Vector and Unit Vector The vector whose magnitude is 0 is called the zero vector. The zero vector has no direction: 0 = 0i + 0j = <0, 0> The vector whose magnitude is 1 is called the unit vector. To find the unit vector in the same direction as a given vector v, divide v by its magnitude Unit Vector = _v_ ||v|| Example: Find the unit vector in the same direction as v = 5i – 12j First find ||v|| = || 5i –12j|| =52 + 122 = 25 + 144 = 169 = 13 Unit Vector = _v_ = 5i – 12j = _5 i - 12 j ||v|| 13 13 13

  22. Using Magnitude and Direction to write a Vector (a,b) v = ||v||cosi + ||v||sinj = < ||v||cos, ||v||sin> ||v||  v = ai + bj Example: Wind is blowing at an angle  = 120 Magnitude ||v|| = 20mph Express the wind’s velocity as a vector v = ||v||cosi + ||v||sinj v = 20cos(120)i + 20sin(120)j v = 20(- ½ )i + 20 (3/2)j v = -10i + 103 j = <-10, 103 >

  23. v = <a, b>w = <c, d> The dot productv • w is defined as follows: Dot Product ac + bd Examples: v = <5, – 2>w = <-3, 4> v · w = 5(-3) + (-2)(4) = -15 – 8 = -23 w · v = (-3)(5) + (4)(-2) = -15 – 8 = -23 v · v. = (5)(5) + (-2)(-2) = 25 + 4 = 29

  24. Another Dot Product example v = <a, b>w = <c, d> ac + bd u = <4, 6> v = <5,– 2>w = <-3, 4> Find: u • (v + w) <4, 6> • <5–2, -3 + 4> <4, 6> • <2, 2> (4)(2) + (6)(2) 8 + 12 20

  25. Dot Product Properties u • v = v • u u • (v + w) = u • v + u • w (u + v) • w = u • w + v • w (ku) • v = k(u • v) = u • kv 0 • u = 0 u • u = |u|2

  26. Finding the angle between vectors with an alternative dot product formula Dot Product Angle Positive Acute 0 Right Negative Obtuse • If v and w are two nonzero vectors and  is the smallest nonnegative angle between them, then Example: Find the angle  between v = <3, -2> and w = <-1, 4> v • u = 2 + 2 = 4 ||v|| = (9 + 4) = 13 ||w||=(1 + 16) = 17 • = cos –1 4/(13)(17) • = cos –1 .2690691176 • = 74.39 degrees

  27. Parallel and Orthogonal Vectors • Two vectors are parallelif the angle between them is 0 or 180° • Two vectors are orthogonalif the angle between them is 90° •  If the dot product of two vectors is 0 then the vectors are orthogonal. Example:v = <3, -2> and w = <3, 2> Are v and w orthogonal? Check the dot product: v • w = (3)(3) + (-2)(2) 9 + -4 5 Since the dot product is NOT zero, these vectors are NOT orthogonal.

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