1. Section 7.1 Oblique Triangles & Law of Sines Section 7.2 Ambiguous Case & Law of Sines Section 7.3 The Law of Cosines Section 7.4 Vectors and the Dot Product Section 7.5 Applications of Vectors Chapter 7 Applications of Trig and Vectors

2. Section 7.1 Oblique Triangles & Law of Sines Congruency and Oblique Triangles Law of Sines Solving using AAS or ASA Triangles

3. Congruency and Oblique Triangles If we use A for angles and S for sides what are all of the three letter combinations you could create? Which of these can we use to prove the triangles are congruent?

4. ASA SAA AAA YES NO Congruence Shortcuts

5. SSS SAS SSA YES NO Congruence Shortcuts

6. Data required for Solving Oblique Triangles 1.One side and two angles (ASA or AAS) 2.Two sides and one angle not included between the two sides (ASS). Yep this one can create more than one triangle 3.Two sides and the angle between them (SAS) 4.Three sides (SSS) 5.Three angles (AAA) Yep this one only creates similar triangles.

7. E C A R Given: AR = ER EC = AC Show /_E = /_A ~ ~ ~ 1. AR = ER 2. EC = AC 3. RC = RC ΔRCE = ΔRCA/_E = /_A ~ ~ ~ ~~ Given Given SSS CPCTC Reflexive

8. E C A R Given: /_E = /_A /_ECR = /_ACR Show AR = ER ~ ~ ~ 2. / ECR = / ACR 3. RC = RC ΔRCE = ΔRCAAR = ER ~ ~ ~ ~ ~ Given Given AAS CPCTC Reflexive 1. /_E = /_A

9. E C A R Given: /_E = /_A /_ERC = /_ARC Show AR = ER ~ ~ ~ 2. / ERC = / ARC 3. RC = RC ΔRCE = ΔRCAAR = ER ~ ~ ~ ~ ~ Given Given ASA CPCTC Reflexive 1. /_E = /_A

10. Law of Sines In any triangle ABC, with sides a, b, and c, =, =, = This can be written in compact form as = = a sin A b sin B a sin A c sin C b sin B c sin C a sin A c sin C b sin B

11. Area of a Triangle In any triangle ABC, the area A is given by any of the following formulas: A = ½bc sin A A = ½ab sin C A = ½ac sin B

12. Section 7.2 Ambiguous Case & Law of Sines Description of the Ambiguous Case Solving SSA Triangles (Case 2) Analyzing Data for Possible Number

13. Ambiguous Case Acute Number of Possible Triangles Sketch Condition Necessary for Case to Hold 0a<h 1a=h 1a>ba>b 2b>a>h

14. Ambiguous Case Number of Possible Triangles Sketch Condition Necessary for Case to Hold 0a<ba<b 1a>b

15. SSA Cases Remember since SSA results in two possible triangles we must check the angles supplement as well. So if we find the angle is 73  then we also have to check 180  – 73  = 107 .

16. Section 7.3 The Law of Cosines Derivation of the Law of Cosines Solving SAS Triangles Case 3 Solving SSS Triangles Case 4 Heron’s Formula for the Area of a Triangle

17. Triangle Side Length Restriction In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side.

18. Law of Cosines

19. In any triangle ABC, with sides a, b, and c, a 2 = b 2 + c 2 – 2bc cos A b 2 = a 2 + c 2 – 2ac cos B c 2 = a 2 + b 2 – 2ab cos C

20. Oblique Triangle Case 1 One side and two angles AAS or ASA 1.Find the remaining angle using the angle sum formula (A+B+C)=180  2.Find the remaining sides using the Law of Sines

21. Oblique Triangle Case 2 Two sides and a non-included angle SSA 1.Find an angle using the Law of Sines 2.Find the remaining angle using the Angle Sum Formula 3.Find the remaining side using the Law of Sines There may be no triangle or two triangles

22. Oblique Triangle Case 3 Two sides and an included angle SAS 1.Find the third side using the Law of Cosines 2.Find the smaller of the two remaining angles using the Law of Sines 3.Find the remaining angle using the angle sum formula

23. Oblique Triangle Case 4 Three sides SSS 1.Find the largest angle using the Law of Cosines 2.Find either remaining angle using the Law of Sines 3.Find the remaining angle using the angle sum formula

24. Heron’s Area Formula If a triangle has sides of lengths a, b, and c, and if the semi-perimeter is s= ½(a+b+c) then the area of the triangle is A =  s(s-a)(s-b)(s-c)

25. Section 7.4 Vectors and the Dot Product Basic Vector Terminology Finding Components and Magnitudes Algebraic Interpretation of Vectors Operations with Vectors Dot Product and the Angle between Vectors

26. Basic Terminology scalars – quantities involving only magnitudes vector quantities – quantities having both magnitude and direction vector – a directed line segment magnitude – length of a vector initial point – vector starting point terminal point – second point through which the vector passes

27. Sum of vectors To find the sum of two vectors A and B: A+B resultant vector or

28. Difference of vectors To find the difference of 2 vectors A and B: A+(-B) resultant vector or

29. To find the product of a real number k and a vector A : kA=A+A+…+A (k times) Example: 3A Scalar Product

30. Magnitude and Direction Angle of a Vector The magnitude of vector u= is given by |u| =  a 2 + b 2 The direction angle  satisfies tan  =b/a, where a ≠ 0.

31. Horizontal and Vertical Components The horizontal and vertical components, respectively, of a vector u having magnitude |u| and direction angle  are given by  = |u| cos  and  =|u| sin 

32. Vector Operations For any real numbers a, b, c, d, and k, + = k · = If a =, then -a = - = + -

33. Unit Vectors i = j = i, j Form for Vectors If v =, then v = ai + bj

34. Dot Product The dot product of the two vectors u = and v = is denoted by u · v, read “u dot v,” and given by u · v = ac + bd

35. Properties of the Dot Product For all vectors u, v, w and real numbers k u · v = v · uu ·( v+w) = u · v + u · w (u+v) · w= u · w + v · w(ku) · v=k(u · v)=u · (kv) 0 · u = 0u · u = |u| 2

36. Geometric Interpretation of Dot Product If  is the angle between the two nonzero vectors u and v, where 0  <  <180 , then u · v = |u||v| cos 

37. Orthogonal Vectors Two nonzero u and v vectors are orthogonal vectors if and only if u · v = 0

38. Section 7.5 Applications of Vectors The Equilibrant Incline Applications Navigation Applications

39. Equilibrant A vector that counterbalances the resultant is called the equilibrant. If u is a vector then –u is the equilibrant. u + -u = 0

40. Equilibrant Force Use the law of Cosines 48 60 48 v -v B A 130à |v| 2 = 48 2 + 60 2 – 2(48)(60)cos(130à) |v| 2 ≈ 9606.5 |v| ≈ 98 newtons The required angle can be found by subtracting angle CAB from 180à. C 9860 sin 130à sin CAB = CAB ≈ 28à so £ = 180à - 28à =152à

41. Inclined Application 20à 50 x sin 20à = |AC|/50 |AC| ≈ 17 pounds of force A C