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

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(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) 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 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

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

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(A) The Sine Law: The Ambiguous Case Let’s work through 2 scenarios of solving for B : 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) 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 Then sin = b sin / a And sin = 2 sin 30 / 3 And sin = 2 sin 30 / 3 So B = 19.5° So B = 19.5°

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(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) 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 Then sin = b sin / a And sin = 3 sin 30 / 2 And sin = 3 sin 30 / 2 So B = 48.6° So B = 48.6° BUT!!!!! …….. there is a minor problem here ….. BUT!!!!! …….. there is a minor problem here …..

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(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} 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) 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: However, if a < b, then there arise three possible cases: (i) if a < b sin A, then no solution (diagram 4) (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) (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) (iii) if a > b sin A, then two triangles are possible (diagram 2)

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(B) Understanding The Ambiguous Case We can visualize the construction of these 4 cases in the following applet: We can visualize the construction of these 4 cases in the following applet: Sine Law - Ambiguous case - applet from AnalyzeMath Sine Law - Ambiguous case - applet from AnalyzeMath Sine Law - Ambiguous case - applet from AnalyzeMath 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) 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)

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(C) Summary We consider the “ambiguous case” under the following condition: 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 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 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 If b sinA < a, then we get 2 possible measures for B (one acute and one obtuse) and thus 2 possible triangles

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(D) Example of Ambiguous Case Back to our previous example: 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) 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° To find B, we take sin -1 (0.75) and get the acute angle of 48.6° Our obtuse angle will then be 180° - 48.6° = 131.4° Our obtuse angle will then be 180° - 48.6° = 131.4° (we can confirm this by simply taking sin(131.4°)) (we can confirm this by simply taking sin(131.4°))

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

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(F) Examples ex. 1. In ABC, A = 42°, a = 10.2 cm and b = 8.5 cm, find the other angles 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 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 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. 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 So second test is a calculation Here a (30) > b sin A (25.66), so there are two cases

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

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