2. FOUR CASES CASE 1: One side and two angles are known (SAA or ASA). CASE 2: Two sides and the angle opposite one of them are known (SSA). CASE 3: Two sides and the included angle are known (SAS). CASE 4: Three sides are known (SSS).

3. CASE 1: ASA or SAA S A A ASA S AA SAA

4. S S A CASE 2: SSA

5. S S A CASE 3: SAS

6. S S S CASE 4: SSS

7. The Law of Sines is used to solve triangles in which Case 1 or 2 holds. That is, the Law of Sines is used to solve SAA, ASA or SSA triangles.

8.   This is used in triangles that are not right triangles. If we know 2 angles and 1 side of the triangle then we can use the law of sines. For example… b 80.4 135 o

9.   The last example shows a case of SAA or AAS  Or ASA. b 80.4 135 o With that set up of ASA or AAS there is one and only one triangle that can be formed.

10. The AMBIGUOUS Case Open to or having several possible meanings or interpretations. Of doubtful or uncertain nature; difficult to comprehend, distinguish, or classify Lacking clearness or definiteness; obscure; indistinct

11. If given SSA there are several situations that can happen.ACb a NO POSSIBILITIES ACb a 1 POSSIBILITY CA b a 2 POSSIBILITIES ACba 1 POSSIBILITY

12. General Steps for solving the ambiguous case You will be given two sides and 1 angle. Use these and the formula for the law of sines to get the second angle Check if the first angle that you found is valid i.e. Do the two angles that you currently have have a sum that is less than 180 Check if there is a second angle that is valid. To do this, find the angle in Quadrant II that has the same sine as the angle that you found in Step 2 (subtract that angle from 180) If the angle in Quadrant II plus the original angle you were given have a sum that is less than 180, you have a second triangle that works.

13. If the given information presents you with a triangle of the orientation SSA, one side can be drawn in two different positions as illustrated below. Because of this we will check for 2 possible triangles. Solve ∆ABC if C b=248.6 A B1B1 a=186.2 B2B2 43.1°

14. b=248.6 A B1B1 a=186.2 B2B2 43.1° From the Law of Sines we can find sinB sin B = 0.91225

15. Using we can find the measure of angle B1, which is approximately 65.82°. To check to see if this angle is valid we start with the angle that was given (43.1° and add 65.82° to ensure that the sum is less than 180°) 43.1°+65.82°=108.92° Thus making the third angle approximately 71.08°.

16. But what about the possibility of a second triangle with angle at B2 because we started with SSA? Take the 65.82° that we found and find its supplement: 180°-65.82°=114.18° (this is done because the sine of an angle is equal to the sine of its supplement, this is due to the fact that sine is positive in quadrants I and II, in hand with the ideas of reference angles.) Now check to see if that angle and the angle we started with of 43.1° has a sum of less than 180°. 114.18° + 43.1° = 157.28°, so with this angle another triangle is possible and its third angle would be 22.72°

17. If either of the angles that I found in the previous example would have been added to 43.1° and the sum of the two angles was greater than 180 °, then that angle would not be a valid angle therefore that triangle would not be able to exist in the geometry that we know (Euclidian), but I would still want to check for a second possibility.

18. 47° 63 ° 5.45” A B C

19. 42°85° 65.85 mm A B C

20. 58° 21° 209 ft A tree on a hillside casts a shadow 209 ft down the hill. If the angle of inclination of the hillside is 21 degrees to the horizontal and the angle of elevation of the sun is 58 degrees, find the height of the tree.

21.  Homework