1. Law of Sines Section 6.1

2. So far we have learned how to solve for only one type of triangle Right Triangles Next, we are going to be solving oblique triangles Any triangle that is not a right triangle

3. In general: C c A a B b

4. To solve an oblique triangle, we must know 3 pieces of information: a)1 Side of the triangle b)Any 2 other components a)Either 2 sides, an angle and a side, and 2 angles

5. AAS ASA SSA SSS SAS C c A a B b Law of Sines

6. If ABC is a triangle with sides a, b, and c, then:

7. ASA or AAS A C B 27.4 102.3 º 28.7 º A = a = c = 49 º 43.06

8. ASA or AAS A C B 27.4 102.3 º 28.7 º A = a = c = 49 º 43.06 55.75

9. Solve the following Triangle: A = 123 º, B = 41º, and a = 10 123 º 41 º 10 C c b C = 16 º

10. 123 º 41 º 10 C c b C = 16 º b = 7.8

11. 123 º 41 º 10 C c b C = 16 º b = 7.8 c = 3.3

12. Solve the following Triangle: A = 60 º, a = 9, and c = 10 60 º 9 C 10 b How is this problem different? B What can we solve for?

13. 60 º 9 C 10 b B C = 74.2 º

14. 60 º 9 C 10 b B C = 74.2 º B = 45.8 º c = 7.5

15. What we covered: Solving right triangles using the Law of Sines when given: 1)Two angles and a side (ASA or AAS) 2)One side and two angles (SSA) Tomorrow we will continue with SSA

16. SSA The Ambiguous Case

17. Yesterday Yesterday we used the Law of Sines to solve problems that had two angles as part of the given information. When we are given SSA, there are 3 possible situations. 1)No such triangle exists 2)One triangle exists 3)Two triangles exist

18. Consider if you are given a, b, and A A a b h Can we solve for h? h = b Sin A If a < h,no such triangle exists

19. Consider if you are given a, b, and A A a b h If a = h,one triangle exists

20. Consider if you are given a, b, and A A a b h If a > h,one triangle exists

21. Consider if you are given a, b, and A A a b If a ≤ b,no such triangle exists

22. Consider if you are given a, b, and A A a b If a > b,one such triangle exists

23. Hint, hint, hint… Assume that there are two triangles unless you are proven otherwise.

24. Two Solutions Solve the following triangle. a = 12, b = 31, A = 20.5 º 20.5 º 31 12

25. 2 Solutions First Triangle B = 64.8 º C = 94.7º c = 34.15 Second Triangle B’ = 180 – B = 115.2 º C ’ = 44.3 º C’ = 23.93

26. Problems with SSA 1)Solve the first triangle (if possible) 2)Subtract the first angle you found from 180 3)Find the next angle knowing the sum of all three angles equals 180 4)Find the missing side using the angle you found in step 3.

27. A = 60 º; a = 9, c = 10 First Triangle C = 74.2 º B = 48.8 º b = 7.5 Second Triangle C’ = 105.8 º B ’ = 14.2 º b ’ = 2.6

28. One Solution Solve the following triangle. What happens when you try to solve for the second triangle? a = 22; b = 12; A = 42 º

29. First Triangle B = 21.4 º C = 116.6º c = 29.4 Second Triangle B’ = 158.6 º C ’ = -20.6 º

30. No Solution Solve the following triangle. a = 15; b = 25; A = 85 º Error → No such triangle