1. Section 8.1 The Law of Sines
Copyright ©2013, 2009, 2006, 2001 Pearson Education, Inc.

2. Objectives Use the law of sines to solve triangles.
Find the area of any triangle given the lengths of two sides and the measure of the included angle.

3. Solving Oblique Triangles
1. AAS: Two angles of a triangle and a side opposite one of them are known.
2. ASA: Two angles of a triangle and the included side are known.

4. Solving Oblique Triangles
3. SSA: Two sides of a triangle and an angle opposite one of them are known. (In this case, there may be no solution,one solution,or two solutions. The latter is known as the ambiguous case.)

5. Solving Oblique Triangles
4. SAS: Two sides of a triangle and the included are known.

6. Solving Oblique Triangles
5. SSS: All three sides of the triangle are known.

7. Law of Sines The Law of Sines applies to the first three situations.
In any triangle ABC, A B C a b c

8. Example In , e = 4.56, E = 43º, and G = 57º. Solve the triangle.
Solution:
Draw the triangle.
We have AAS.

9. Example (cont) Find F: F = 180º – (43º + 57º) = 80º
Use law of sines to find the other two sides.

10. Example
We have solved the triangle.

11. Solving Triangles SSA
When two sides of a triangle and an angle opposite one of them are known, the law of sines can be used to solve the triangle. There are various possibilities as show in the following 8 cases.
Angle B is acute
Case 1: No solution
b < c; side b is too short to reach the base. No triangle formed.

12. Solving Triangles SSA
Angle B is acute Case 2: One solution b < c; side b just reaches the base and is perpendicular to it.
Angle B is acute
Case 3: Two solutions
b < c; an arc of radius b meets the base at two points. This is called the ambiguous case.

13. Solving Triangles SSA
Angle B is acute Case 4: One solution b = c; an arc of radius b meets the base at just one point other than B.
Angle B is acute
Case 5: One solution
b > c; an arc of radius b meets the base at just one point.

14. Solving Triangles SSA
Angle B is obtuse Case 6: No solution b < c; side b is too short to reach the base. No triangle formed.
Angle B is obtuse
Case 7: No solution
b = c; an arc of radius b meets the base only point B. No triangle is formed.

15. Solving Triangles SSA
Angle B is obtuse Case 8: One solution b > c; an arc of radius b meets the base at just one point.
The eight cases lead us the three possibilities in the SSA situation: no solution, one solution, or two solutions.

16. Example In , b = 15, c = 20, and B = 29º. Solve the triangle. Solution
Draw a triangle.
Find C.

17. Example (cont)
There are two angles less than 180º with a sine of : 40º and 140º. So there are two possible solutions.
Possible Solution I If C = 40º, then
A = 180º – (29º + 40º) = 111º

18. Example (cont) Then we find a:
These measures make a triangle as shown. Thus we have a solution.

19. Example (cont)
Possible Solution II If C = 140º, then A = 180º – (29º + 140º) = 11º
Then we find a:
These measures make a triangle as shown. Thus we have a solution.

20. The Area of a Triangle
The area of any is one half the product of he lengths of two sides and the sine of the included angle:

21. Example
A university landscaping architecture department is designing a garden for a triangular area in a dormitory complex. Two sides of the garden, formed by the sidewalks in front of buildings A and B, measure 172 ft
and 186 ft, respectively, and together form a 53º angle. The third side of the garden, formed by the sidewalk along Crossroads Avenue, measures 160 ft. What is the area of the garden to the nearest square foot?

22. Example (cont) Solution: Use the area formula.
The area of the garden is approximately 12,775 ft2.