Presentation on theme: "Ambiguous Case Triangles"— Presentation transcript:
1 Ambiguous Case Triangles Can given numbers make a triangle? Can a different triangle be formed with the same information?
2 Conditions for Unique Triangles AASASAtwo angles must sum to less than 180ºtwo angles must sum to less than 180ºSSSSAStwo shortest sides are longer than the third sideAny set of data that fits these conditions will result in one unique triangle.
3 Ambiguous Triangle Case (aka the ‘bad’ word) SSAAabThis diagram is deceiving -- side-side-angle data may result in two different triangles.Side a is given but it might be possible to ‘swing’ it to either of two positions depending on the other given values.An acute or an obtuse triangle may be possible.
4 Determining the Number of Possible Triangles 3.4 The Ambiguous Case Of The Law Of SinesIf two sides and one angle opposite to one of them are given, there can be complications in solving triangles. In fact, from the given data, we may determine more than one triangle or perhaps no triangles at all.Let the angle be C and sides be c and b of triangle ABC given thenHere, R.H.S. is completely known and hence Ð B can be found out.Also Ð C is given \ ÐA = (B + C). Thus triangle ABC is solved.But when sin B has R. H.S. with such values that the triangle can't be com pleted.Observe,(I) When Ð C < 900 (acute angle)(a) If c < b sin C then 1 i.e. sin B > 1 which is impossible (as sine c ratio never exceeds 1)(b) c = b sin C then = 1 i.e. sin B = 1 \ Ð B = 900 i.e triangle ABC is c right triangle.(c) If c > b sin C then < 1 i.e. sin B < 1. But within the range 00 to 1800 ; there are two values of angles for a sine ratio as sin q = sin (180 - q), of which one is acute and the other is an obtuse. e.g. sin 300 = sin 1500But both of these may not be always admissible.If ÐB < 900 (acute), the triangle ABC is possible as c > b then Ð C > Ð B. But if Ð B > 900 (obtuse) then Ð C will be also obtuse ; this is impossible as there can't be two obtuse angles in a triangle.If c < b and Ð C is acute then both values of B are admissible . Naturally there will be two values of Ð A and hence two values of a. Hence there are two triangle possible.(II) When Ð C < 900 (obtuse angle).Here (1) If c < b then Ð C < Ð B but then B will also become an obtuse angle. It is also impossible.(2) If c = b then Ð B = Ð C. Hence Ð C is obtuse makes Ð B obtuse too which again impossible.(3) If c > b then Ð C > Ð B. But again there are two possibilities (i) Ð B is acute (ii) and Ð B is obtuse.If Ð B is acute the triangle is possible and when Ð B is obtuse the triangle is impossible. For the obtained values of the elements when there is ambiguity (i.e. we are unable to draw such a triangle) to determine the triangle ; it is called an 'ambiguous case'.The Textbook Method1) Use Law of SinesThe easy way2) Sum of the angles in a triangle = 180º
5 Both values of C are possible, so 2 triangles are possible Example (2 triangles)Given informationSet up Law of SinesSolve for sin BmA = 17ºa = 5.8b = 14.3mB 46ºFind mB in quadrant IFind mB in quadrant IImB 180 – 46 = 134ºmC (180 – 17 – 46) 117ºFind mCmC (180 – 17 – 134) 29ºFind mCBoth values of C are possible, so 2 triangles are possible
6 Only one value of C is possible, so only 1 triangle is possible Example (1 triangle)Given informationSet up Law of SinesSolve for sin BmA = 58ºa = 20b = 10mB 25ºFind mB in quadrant IFind mB in quadrant IImB 180 – 25 = 155ºmC (180 – 58 – 25) 97ºFind mCmC (180 – 58 – 155) -33ºFind mCOnly one value of C is possible, so only 1 triangle is possible
7 No value of B is possible, so no triangles are possible Example (0 triangles)Given informationSet up Law of SinesSolve for sin BmA = 71ºa = 12b = 17No value of B is possible, so no triangles are possible
8 Law of Sines Method 1) Use Law of Sines to find angle B -If there is no value of B (for example, sin B = 2), then there are no trianglesRemember, sin x is positive in both quadrant I and II2) Determine value of B in quadrant II(i.e. 180 – quadrant I value)3) Figure out the missing angle C for both values of angle B by subtracting angles A and B from 1804) If it is possible to find angle C for-both values of B, then there are 2 triangles-only the quadrant I value of B, then only 1 triangle is possible