# TF.01.3 – Sine Law: The Ambiguous Case

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TF.01.3 – Sine Law: The Ambiguous Case
MCR3U - Santowski

(A) The Sine Law: The Ambiguous Case
At this stage of solving triangles, we can work with the Sine Law if we know both an angle and its opposite side (if not, we have to try the Cosine Law) So, let’s start with ABC, where A is acute, and we know the measure of sides a and b and we wish to solve for B or for side c

(A) The Sine Law: The Ambiguous Case
So to solve for B: a/sin  = b/sin  a sin  = b sin  sin  = (b sin ) ÷ a sin  = (b sin )/a So as long as (b sin )/a > 1, then we can solve for B

(A) The Sine Law: The Ambiguous Case
Let’s work through 2 scenarios of solving for B : Let A = 30°, a = 3 and b = 2 (so the longer of the two given sides is opposite the given angle) Then sin  = b sin  / a And sin  = 2 sin 30 / 3 So B = 19.5°

(A) The Sine Law: The Ambiguous Case
In our second look, let’s change the measures of a and b, so that a = 2 and b = 3 (so now the shorter of the two given sides is opposite the given angle) Then sin  = b sin  / a And sin  = 3 sin 30 / 2 So B = 48.6° BUT!!!!! …….. there is a minor problem here …..

(B) Understanding The Ambiguous Case
Construct four triangles ABC such that  A = 30°, b = 2.0 and a = {2.3; 1.5; 1.0; 0.8} So we make the observation that if a > b, then we have only one possible triangle that can be constructed (diagram 1) However, if a < b, then there arise three possible cases: (i) if a < b sin A, then no solution (diagram 4) (ii) if a = b sin A, then one right triangle where b sin A is the altitude from AB from C (diagram 3) (iii) if a > b sin A, then two triangles are possible (diagram 2)

(B) Understanding The Ambiguous Case
We can visualize the construction of these 4 cases in the following applet: Sine Law - Ambiguous case - applet from AnalyzeMath And so we see that under special situations (a < b), we can, at times, construct two unique triangles (one where B is acute and one where  B is obtuse) from the given information (SSA)

(C) Summary We consider the “ambiguous case” under the following condition: We are given two sides (a & b) and one angle (A), wherein the side opposite the given angle is the SHORTER of the two sides Then we do a mathematical calculation  we take the product of b sinA (which represents the altitude in the triangle) and compare it to the measure of side a If b sinA < a, then we get 2 possible measures for  B (one acute and one obtuse) and thus 2 possible triangles

(D) Example of Ambiguous Case
Back to our previous example: Let A = 30°, a = 2 and b = 3 (so now the shorter of the two given sides is opposite the given angle) To find B, we take sin-1(0.75) and get the acute angle of 48.6° Our obtuse angle will then be 180° ° = 131.4° (we can confirm this by simply taking sin(131.4°))

(E) Internet Links Now watch an on-line teaching video, showing you examples of the sine law and this ambiguous case An on-line teaching video showing examples of "The Ambiguous Case" from MathTV.com

(F) Examples ex. 1. In ABC, A = 42°, a = 10.2 cm and b = 8.5 cm, find the other angles First test side opposite the given angle is longer, so no need to consider the ambiguous case  i.e. a > b  therefore only one solution ex. 2. Solve ABC if A = 37.7, a = 30 cm, b = 42 cm First test  side opposite the given angle is shorter, so we need to consider the possibility of the “ambiguous case”  a < b  so there are either 0,1,2 possibilities. So second test is a calculation  Here a (30) > b sin A (25.66), so there are two cases

(G) Homework Nelson text, pg511, Q5,6,8