1. Ambiguous Case of the Law of Sines Section 5.7

2. We get the ambiguous case when we are given SSA. Given a, b and A, find B. In calculator, inverse key gives only the 1 st quadrant angle, but sin is positive in the 2 nd quadrant also.

3. Three Cases  third side is too short sinB > 1 (calculator gives an ERROR!) therefore, NO TRIANGLE!  third side is just long enough to hit sinB = 1, therefore B = 90 o triangle is a right triangle, so we can use SOHCAHTOA  third side is longer and hits in two points sinB < 1 we could have two triangles and must solve for BOTH

4. What does all that mean? Let’s take a look …

5. If we think we have two triangles …  Calculator gives us the first quadrant angle  Sin is also positive in the 2 nd quadrant, so we need to that angle find the supplement of the angle given by the calculator  Determine whether there is a 2 nd triangle. By adding given angle and supplementary angle. If sum > 180, no 2 nd triangle. Proceed with calculator given angle. If sum < 180, we do have a 2 nd triangle. Proceed to solve BOTH triangles.

6. Solve the triangle: a = 22, b = 12, A = 42 o

7. Solve the triangle: a = 15, b = 25, A = 85 o

8. Solve the triangle: a = 12, b = 31, A = 20.5 o

9. Assignment  pg 324 #18 – 29