70 likes | 162 Views
Aim: How do we handle the ambiguous case?. Do Now: In ∆ ABC , m A = 30° , a = 6 and c = 10. Find C to the nearest degree. We can use the Law of Sine to answer the problem. HW: p.574 # 6b,8b,10b,14b,16,18. Notice that sin C is positive 56 ° which can be in
E N D
Aim: How do we handle the ambiguous case? Do Now: In ∆ABC, mA = 30° , a = 6 and c = 10. Find Cto the nearest degree. We can use the Law of Sine to answer the problem. HW: p.574 # 6b,8b,10b,14b,16,18
Notice that sin C is positive 56° which can be in quadrant I or quadrant II. If angle C is in quadrant I then it is 56°. If56°is in quadrant II then it is 180° – 56°= 124° Which angle measurement is correct for angle C? We call this situation as ambiguous case that means there are different choices for C
How do we deal with this ambiguous case? We have to use the basic idea about the sum of the interior angles of a triangle to figure out. We are given A = 30 If C = 56°, then B = 180° – (56 + 30) = 94° If C = 124°, then B = 180° – (124° + 30) = 26° Both angle measurements satisfy the basic requirement of interior angles of a triangle. Therefore, we can say that there are two different triangles can be made with the given condition.
Let’s see another example: and c = 12 In a) Find mC b) How many triangles can be drawn?
or If then If Two interior angles of the ∆ABC are already over 180° Therefore, mC =156° is invalid. The answer for a) is 24° only Therefore, there is only one triangle can be made with the given condition.
In ΔABC, b = 18, c = 10 and B = 70. How many triangles can be constructed? B 70 11 A 10 C C undefined, therefore no triangle can be constructed
Application: • If side a = 16, side b = 20, mB = 30°, how many • distinct triangles can be constructed? 1 triangle 2. If mA = 68°, side a = 10 and side b = 24, how many distinct triangles can be constructed? 0 triangle 3. If side a = 18, side b = 10 and mC = 70°, how many distinct triangles can be constructed? 2 triangles