2. Opening Questions Please solve these… A 7” 9” #1 Find the measure of ےA 30° 10” x #2 Find x

3. Trigonometry Working with Oblique Triangles: Law of Sines

4. Introduction Given a right triangle, you should feel comfortable using the three basic trig functions to determine additional information about the triangle.

5. Introduction Depending upon the information given, you could either determine the size of an angle... A 7” 9” opposite hypotenuse

6. Introduction …or determine the length of a side. 30° 10” x hypotenuse opposite

7. Introduction But what happens when you must work with an oblique triangle (one without a right angle)? Sometimes you can split-up an oblique triangle into right triangles...

8. Introduction …but often it is better to use a more appropriate “tool” for the job. One of those tools for working with oblique triangles is called the Law of Sines.

9. Law of Sines The Law of Sines states that in any triangle, the sides are proportional to the sines of the opposite angles. AB C c a b

10. Law of Sines Side a is proportional to the sine of angle A... AB C c a b

11. Law of Sines …side b is proportional to the sine of angle B... AB C c ab

12. Law of Sines …and side c is proportional to the sine of angle C. AB C c a b

13. 47° 63° 5.45” x Law of Sines This known side is opposite one of the given angles. There are two kinds of situations where you can use the Law of Sines. –Situation 1: When you know two angles, the length of a side opposite, and you want to determine the length of another side.

14. 47° 63° 5.45” x Law of Sines There are two kinds of situations where you can use the Law of Sines. –Situation 1: When you know two angles, the length of a side opposite, and you want to determine the length of another side. This side is opposite the other given angle.

15. 42° 85.5 mm 70 mm A Law of Sines There are two kinds of situations where you can use the Law of Sines: This angle is opposite one of the given sides. –Situation 2: You know the length of two sides, the size of an angle opposite, and you want to determine the size of another angle.

16. 42° 85.5 mm 70 mm A Law of Sines There are two kinds of situations where you can use the Law of Sines: –Situation 2: You know the length of two sides, the size of an angle opposite, and you want to determine the size of another angle. This “unknown” angle is opposite the other side.

18. Law of Sines Example 1 Given the diagram below, determine the length of side x. 47° 63° 5.45” x

19. 47° 63° 5.45” x Law of Sines Example 1 What do we know about this problem? First of all, it is an oblique triangle. Second, we note that two angles are known, and one of the sides opposite.

20. Law of Sines Example 1 That’s enough info to verify that using the Law of Sines will allow us to determine the length of x. 47° 63° 5.45” x

21. Law of Sines Example 1 To solve, set up a proportion. Remember that the sides are proportional to the sines of the opposite angles. 47° 63° 5.45” x

22. Law of Sines Example 1 Start by pairing the 63° angle and the 5.45” side together since they are opposite one another. 47° 63° 5.45” x

23. Law of Sines Example 1 The unknown side x is opposite the 47° angle. Pair these up to complete the proportion. 47° 63° 5.45” x

24. Law of Sines Example 1 Solve the proportion by cross-multiplying. 47° 63° 5.45” x Multiply on this diagonal first. 5.45 x sin47° = 3.99

25. Law of Sines Example 1 Solve the proportion by cross-multiplying. 47° 63° 5.45” x Next, divide 3.99 by sin63° 3.99 ÷ sin63° = 4.47 (this is the length of x)

26. Law of Sines Example 1 By using the Law of Sines, we know the length of side x is 4.47 inches. 47° 63° 5.45” 4.47”

28. Law of Sines Example 2 Given the diagram below, determine the length of side x. 42°85° 65.85 mm x

29. Law of Sines Example 2 Before you jump in, be sure you know what you are dealing with. You are working with an oblique triangle... …and you know two angles and a side opposite one of those angles. 42°85° 65.85 mm x

30. Law of Sines Example 2 That means using the Law of Sines will allow you to solve for x. 42°85° 65.85 mm x

31. Law of Sines Example 2 Set-up a proportion, starting with the 65.85 mm side and the 85° angle since they are opposite one another. 42°85° 65.85 mm x

32. Law of Sines Example 2 Then complete the proportion by making another ratio using side x and the 42° angle. 42°85° 65.85 mm x

33. Law of Sines Example 2 Solve the proportion. 42°85° 65.85 mm x Multiply on this diagonal first. 65.85 x sin42° = 44.1

34. Law of Sines Example 2 Solve the proportion. 42°85° 65.85 mm x Next, divide 44.1 by sin85° 44.1 ÷ sin85° = 44.2 (this is the length of x)

35. Law of Sines Example 2 Using the Law of Sines on this problem gives you an answer of 44.2 mm. 42°85° 65.85 mm 44.2 mm

37. Law of Sines Example 3 Try this one on your own. Set-up a proportion and solve for x. –Then click to see the answer. 88° 57° 9.25 cm x

38. Law of Sines Example 3 How did it turn out? 88° 57° 9.25 cm x x = 11.02 cm

39. Law of Sines Recall that the other scenario where you can use the Law of Sines is when you know the lengths of two sides and the size of an angle opposite on of those sides. 42° 85.5 mm 70 mm

41. Law of Sines Example 4 Given the diagram below, determine the size of angle A. 42° 85.5 mm 70 mm A

42. Law of Sines Example 4 Once again, set-up a proportion. Start by pairing-up the 70 mm side and the 42° angle. 42° 85.5 mm 70 mm A

43. Law of Sines Example 4 Complete the proportion by putting the 85.5 mm side and angle A together. 42° 85.5 mm 70 mm A

44. Solve the following Triangle: B u t W a i t ! T h e r e a r e t w o o p t i o n s !

45. When given two sides & an angle you can have: 1.1 distinct triangle 2.2 Distinct triangles 3.No triangle that will fit the data When give two angles & a side, there is always exactly one distinct triangle that can be formed.

46. Here’s the trick to knowing which case you have… I f t h e s i d e t h a t i s p a i r e d w i t h t h e k n o w n a n g l e i s b i g g e r t h a n t h e o t h e r s i d e, t h e n t h e r e i s e x a c t l y o n e s o l u t i o n. I f t h e s i d e t h a t i s p a i r e d w i t h t h e k n o w n a n g l e i s s m a l l e r t h a n t h e o t h e r s i d e, t h e n t h e r e i s e i t h e r t w o s o l u t i o n s o r n o s o l u t i o n s – t o f i g u r e o u t w h i c h i t i s, p r o c e e d t o s o l v e i t. I f t h e r e ’ s n o s o l u t i o n, i t w i l l b e c o m e o b v i o u s !

47. Try this one…

49. Now, try this one…