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Ch6.1A – Law of Sines - helps us solve oblique triangles. (No rt angles) (can be flipped over)

Ch6.1A – Law of Sines - helps us solve oblique triangles. (No rt angles) (can be flipped over) C b a Proof: A B C b A B. c. a. c.

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Ch6.1A – Law of Sines - helps us solve oblique triangles. (No rt angles) (can be flipped over)

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  1. Ch6.1A – Law of Sines - helps us solve oblique triangles. (No rt angles) (can be flipped over) C b a Proof: A B C b A B c a c

  2. Ex1) Find the remaining angle and sides 102.3˚ 27.4 ft 28.7˚

  3. Ex2) A pole tilts towards the sun at 8˚ off the vertical, casting a 22 ft shadow. If the angle of elevation of the sun is 43˚, how tall is the pole? 8˚ 43˚ 22 ft

  4. SSA can have 3 possibilities: 1 solution, 2 solutions, or no solution! Ex3) Find the remaining side and angle 22in 12in 42˚

  5. Ex4) Show that no triangle works for a=15, b=25, A=85˚. Ex5) Find 2 triangles for a=12, b=31, A=20.5˚. Ch6.1A p466 1–17odd 25 12 85˚ 31 12 20.5˚

  6. Ch6.1A p466 1–17odd

  7. Ch6.1A p466 1–17odd

  8. Ch6.1B – Area of an Oblique Triangle Area = ½.b.c.sinA = ½.a.b.sinC = ½.a.c.sinB HW#35) Find area for a=4, b=6, C=120˚. 6 120˚ 4

  9. HW#21) A flagpole is located on a 12˚ slope. It casts a shadow 16m up the slope. Find the height of the pole.

  10. HW#23) A 10m pole casts a shadow 17m down a slope. What is the angle the pole makes with the down-slope?

  11. HW#24) A plane flies 500km at a bearing of N44˚W from B to C. the plane then flies 720km from C to A. Find angle C. 720km 500km 44˚ A B C

  12. HW#25) A bridge is to be built across a lake from B to C. A surveyor measures angle B to be 65˚ then walks 100m to A and measures its angle to be 46˚. How long will the bridge be? Ch6.1B p466 21-26,29,35,37,39 A 46˚ B 65˚ C

  13. Ch6.1B p466 21-26,29,35,37,39

  14. Ch6.1B p466 21-26,29,35,37,39

  15. Ch6.1B p466 21-26,29,35,37,39

  16. Ch6.1B p466 21-26,29,35,37,39

  17. Ch6.2A – Law of Cosines - solves oblique triangles for SSS or SAS. - always do longest side/biggest angle first! Ex1) Find the 3 angles of the triangle: B 14 8 A C 19

  18. Ex2) The pitcher’s mound on a softball field is 46ft from home plate and the distance between bases is 60ft. How far is the pitcher’s mound from 1st base?

  19. Ch6.2B p474 1-9odd,19,23(find angles),25

  20. Ch6.2B Heron’s Area Formula Ex3) A trip travels 60mi due east, then adjusts its course 15˚ northward and travels 80mi. How far is it from its starting point? 80mi 15˚ 60mi

  21. Heron’s Area Formula Ex4) Find the area of a triangle having sides a = 43m, b = 53m, c = 72m. Ch6.2B p474 18-26even, 37,39 (22,26 in class)

  22. Ch6.2B p474 18-26even, 37,39 (22,26 in class)

  23. Ch6.2B p474 18-26even, 37,39 (22,26 in class)

  24. Ch6.3A Vectors u

  25. Ch6.3A Vectors v u Ex1) Let u be a vector from P (0,0) to Q (3,2) Double uprights Let v be a vector from R (1,2) to S (4,4) means magnitude Show that ||u|| = ||v|| (length).

  26. Vector Components If a vector goes from point P (p1,p2) to point Q (q1,q2) then its components <v1,v2> = <q1 – p1, q2 – p2> The length of vector v is given by Ex2) Find the component form and length of vector v that has the initial point (4,-7) and terminal point (-1,5).

  27. Vector Operations Scalar multiplication - multiply vector u by a scalar k will make it longer or shorter k.u = k<u1,u2> = <ku1,ku2> (If k is (+) points same direction, if k is (–) points opposite.) Vector Addition u+ v = <u1+ v1, u2 + v2> Ex3) Let v = <-2,5> and w = <3,4>, find: a) 2v b) w + v c) w – v d) v + 2w Ch6.3A p486 1-19odd

  28. Ch6.3A p486 1-19odd

  29. Ch6.3A p486 1-19odd

  30. Ch6.3A p486 1-19odd

  31. Ch6.3A p486 1-19odd

  32. Ch6.3B Unit Vectors u <-2,5>

  33. Ch6.3B Unit Vectors v <-2,5> A unit vector has a length of 1 unit and points in the direction of the original vector. Formula: unit vector u = Ex1) Find the unit vector in the direction of <-2,5>

  34. The standard unit vectors are <1,0> and <0,1>. They are called i and j. i= <1,0> j = <0,1>. Use them to represent a vector in its component form: v = <v1,v2> = v1<1,0> + v2<0,1> = v1i + v2j Ex2) Let u be the vector from (2,-5) to (-1,3). Write it as a combo of the unit vectors I and j.

  35. Ex3) Let u= -3i + 8j and v = 2i – j Find 2u – 3v Ch6.3B p486 21-37odd

  36. Ch6.3B p486 21-37odd

  37. Ch6.3B p486 21-37odd

  38. Ch6.3C Direction Angles u is a unit vector u at an angle θ. θ

  39. Ch6.3C Direction Angles u is a unit vector u at an angle θ. θ u = <cosθ, sinθ> u = (cosθ)i + (sinθ)j

  40. Ch6.3C Direction Angles v u is a unit vector u at an angle θ. θ u = <cosθ, sinθ> u = (cosθ)i + (sinθ)j Any vector v that points the same direction as u can be represented by: v = ||v||<cosθ, sinθ> v = ||v||(cosθ)i + ||v||(sinθ)j

  41. Ch6.3C Direction Angles v u is a unit vector u at an angle θ. θ u = <cosθ, sinθ> u = (cosθ)i + (sinθ)j Any vector v that points the same direction as u can be represented by: v = ||v||<cosθ, sinθ> v = ||v||(cosθ)i + ||v||(sinθ)j To find the angle, given the vector v = ai + bj

  42. Ex1) Give the direction angle for : a) u = 3i + 3j b) v = 3i – 4j Ex2) Find the component form of the vector that represents the velocity of an airplane descending at 100mph at an angle of 30˚. v = ||v||(cosθ)i + ||v||(sinθ)j

  43. Ex3) An airplane is flying north at 500mph, encounters a wind blowing northeast (45˚) at 70mph. What is the resulting speed and direction of the airplane? Ch6.3C p487 39-55odd,63,65

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