The document discusses the Law of Sines, which can be used to solve oblique triangles (non-right triangles) when given two angles and the side between them (AAS), two sides and the angle opposite one of the sides (SSA), or one angle, one adjacent side, and one opposite side. It provides examples of using the Law of Sines to solve triangles in these cases. It also discusses a special case where the opposite side could form two possible triangles, requiring consideration of both solutions.

1. Law of Sines

4. Playing with a the triangle Let’s drop an altitude and call it h . h If we think of h as being opposite to both A and B, then Let’s solve both for h . This means A B C a b c

5. If I were to drop an altitude to side a , I could come up with Putting it all together gives us the Law of Sines. You can also use it upside-down. A B C a b c

8. Example 1 (AAS) Once I have 2 angles, I can find the missing angle by subtracting from 180. C=180 – 45 – 50 = 85 I’m given both pieces for sinA/a and part of sinB/b, so we start there. Cross multiply and divide to get A B C a b c 45 50 =30 85

9. We’ll repeat the process to find side c. Remember to avoid rounded values when computing. Cross multiply and divide to get We’re done when we know all 3 sides and all 3 angles. 45 50 =30 85 A B C a b c

10. Example 2 ASA Let A = 35, B = 10, and c = 45 35 10 45 Since we can’t start one of the fractions, we’ll start by finding C. C = 180 – 35 – 10 = 135 135 Since the angles were exact, this isn’t a rounded value. We use sinC/c as our starting fraction. Cross multiply and divide Using your calculator 36.5 11.1 A B C a b c

11. Example 3 SSA A B C a b c Let A = 40, b = 10, and a = 7 We have enough information to use sinA/a as our starting fraction and can go to work on B We cross multiply and divide to get We have to get rid of the sin function to find the answer for angle B. To do this, we apply the inverse operation. With a calculator 40 10 7 66.7

12. We need something to start the fraction sinC/c. Since it is critical that the angles total 180, we will find C by subtracting from 180. C=180 – 40 – 66.7 = 73.3 73.3 Now we use our starting fraction, sinA/a with sinC/c. Cross multiply and divide to finish. 10.4 40 10 7 66.7 A B C a b c

13. A Pain in the A ngle S ide S ide Let’s consider the case where we have an angle, an adjacent side, and an opposite side. For example, I have angle A, side b, and side a. Sometimes a is too short to reach. You attempt to work this problem like example 3. Your calculator will give you an error message and catch the error. Sometimes a is just right. It reaches with a right angle. You work this problem like example 3 and will never know there might have been a problem. A b a A b a

14. Sometimes a is so long it only reaches one way. This problem also works just like example 3. You’ll never know this might have been difficult. Sometimes a is just the right length that it can form 2 different triangles. Following example 3 solves the outer triangle. You have to be on the look-out to catch the second triangle. If you don’t get an error and a < b, then there will be either a right angle or 2 two triangles. A b a No triangle on this side a a A b

15. Contemplating Triangles A b a a Outer triangle Inner triangle C c B B B’ c’ Solving like example 3 will give you the purple (outer) triangle. Since the 2 red lines (a) are equal, the 2 base angles (B) are also equal. Since B and B’ form a straight line, B’ = 180 – B. Use B’ to compute C’. Subtract A and B’ from 180 to find C’ C’

16. Another Perspective on Starting the Second Triangle a a If the 2 sides are equal, the 2 angles are equal. B B Since supplementary angles total 180, B’ = 180 – B. B’ = 180 - B A b c’ C’ Since A, b, and a are given, once you know B’ you can finish the triangle on the left.

17. An Example of a Pain in the A ngle S ide S ide Let A = 40, b = 10, and a = 9. Angle side side A B C a = 9 We will proceed like example 3. We have enough information for A and a start for B Cross multiply and divide to get c 40 b = 10

18. To get to angle B, you must unlock sin using the inverse. A B C a = 9 45.6 Once you know 2 angles, find the third by subtracting from 180. C = 180 – 40 – 45.6 = 94.4 94.4 c We’re ready to look for side c. Remember to use the starter fraction (A) Cross multiply and divide. Then use a calculator. =14.0 40 b = 10

19. Finding the Second Triangle A 40 b= 10 a = 9 B = 45.6 B’ Start by finding B’ = 180 - B B’ = 180 – 45.6 = 134.4 Now solve this triangle. A B’ C’ b= 10 a = 9 40 134.4 c’

20. Start by finding C’ = 180 – 40 – 134.4 C’ = 5.6 5.6 Finish by finding side c, using your starter fraction (A) =1.4 A B’ C’ b= 10 a = 9 40 134.4 c

21. A New Way to Find Area We all know that A = ½ bh. And a few slides back we found this. It looks like the base is c . Making a little substitution gives me If we drop the altitude in a different direction, we can add It’s always the 2 sides and the included angle. A B C a b c