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 a b A B c

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

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

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

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

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

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

10. C
C = 16º 10
b = 7.8 b
123º
41º c

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

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

13. C
C = 74.2º 9 b
60º B 10

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

15. What we covered:
Solving right triangles using the Law of Sines when given:
Two angles and a side (ASA or AAS)
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.
No such triangle exists
One triangle exists
Two triangles exist

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

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

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

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

22. Consider if you are given a, b, and A
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 31 12 20.5º Solve the following triangle.
a = 12, b = 31, A = 20.5º 31 12
20.5º

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

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

27. A = 60º; a = 9, c = 10 C = 74.2º B = 48.8º b = 7.5 C’ = 105.8º
First Triangle
Second Triangle
C = 74.2º
B = 48.8º
b = 7.5
C’ = 105.8º
B’ = 14.2º
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 = 42º

29. a = 22; b = 12; A = 42º B = 21.4º C = 116.6º c = 29.4 B’ = 158.6º
First Triangle
Second Triangle
B = 21.4º
C = 116.6º
c = 29.4
B’ = 158.6º
C’ = -20.6º

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

31. Law of Sines
Section 6.1

32. Warm Up Solve the following triangles (if possible)
A = 58º; a = 20; c = 10
B = 78º; b = 207; c = 210
A = 62º; a = 10; b = 12

33. A = 58º; a = 20; c = 10
C = 25º
B = 97º
b = 23.4

34. B = 78º; b = 207; c = 210
C = 82.9º
A = 19.1º
b = 69.2

35. B = 78º; b = 207, c = 210 C = 82.9º A = 19.1º a = 69.2 C’ = 97.1º
First Triangle
Second Triangle
C = 82.9º
A = 19.1º
a = 69.2
C’ = 97.1º
A’ = 4.9º
a ’ = 18.1

36. A = 62º; a = 10; b = 12
No such triangle

37. Area of an oblique triangle

38. What is the area of a triangle?b
b = c h
h = b Sin A A

39. Area of an Oblique Triangle
The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle.
i.e.: Area = ½ bc Sin A
= ½ ac Sin b
= ½ ab Sin B

40. Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102º.
Area = ½ (Side) (Side) Sine (angle)
Area = ½ (90) (52) Sine (102º)
Area = 1,189 m²

41. Find the area of the triangle:
A = 35º, a = 14.7, b = 14.7

42. The course for a boat race starts at a point A and proceeds in the direction S 52º W to point B, then in the direction of S 40º E to point C, and finally back to A. Point C lies 8 miles directly south of point A. How far was the race? A 52 B 40 C

43. An airplane left an airport and flew east for 169 miles
An airplane left an airport and flew east for 169 miles. Then it turned northward to N 32º E. When it was 264 miles from the airport, there was an engine problem and it turned to take the shortest route back to the airport. Find θ, the angle through which the airplane turned.

44. A farmer has a triangular field with sides that are 450 feet, 900 feet, and an included angle of 30º. He wants to apply fall fertilizer to the field. If it takes one 40 pound bag of fertilizer to cover 6,000 square feet, how many bags does he need to cover the entire field?