Implementing the Law of Sines to solve SAS triangles

Transcription

1 Implementing the Law of Sines to solve SAS triangles June 8, 009 Konstantine Zelator Dept. of Math an Computer Siene Rhoe Islan College 600 Mount Pleasant Avenue Proviene, RI 0908 U.S.A. kzelator@ri.eu konstantine_zelator@yahoo.om. Introution Most trigonometry or prealulus texts offer a fairy rief treatment of the sujet of solving a triangle; while only a few ontain a more extensive analysis of the sujet matter. By solving a triangle, we refer to the prolem of etermining all three sielengths an interior triangle angle on given information on some of the triangle s angles an/or sielengths. Depening Effetive August, 009 an for the aaemi year : Konstantine Zelator Department of Mathematis 0 Thakeray Hall 9 University Plae University of Pittsurgh Pittsurgh, PA kzet59@pitt.eu on the speifi information, a unique triangle may e forme, more than one triangle, or no triangle at all. Roert Blitzer s ook, Prealulus, is among those ooks that ontain a typial treatment of this sujet (see []). Now, the ase SAS refers to the situation wherein two sielengths are given, as well as the egree measure of the angle ontaine etween the two sies. As we know from Euliean geometry, suh a triangle is uniquely etermine; meaning that any two triangles onstrute with these speifiations must e ongruent. It appears that not only Blitzer s ook, ut universally, in any ook or text with the SAS ase, the Law of Cosines is employe. For example, let us say that the lengths a an are given; as well as the egree measure of the angle BCA (see Figure ). Then, y using the Law of Cosines, one alulates the value of, the length of BA ; an y using the Page of 9

2 Law of Sines, one etermines the values of sin ϕ an sin θ. By the use of the inverse funtion on a alulator if neessary, one etermines the egree measures ϕ an θ. B C a ϕ θ A Figure In this work, we present an alternative approah to solving an SAS triangle. An approah whose key ingreient is the Law of Sines. Law of Sines a sinϕ sin () Basi trigonometri ientities The following ientities are wiely very well known an an e foun in any stanar trigonometry or prealulus text. For any angle egree measures α an β, os sin ( α + β ) osα os β sinα sin β ( α + β ) sinα os β + osα sin β () For any egree measures α an β, α + β α β sinα + sin β sin os () α β α + β sinα sin β sin os For any angle egree measure α not of the form 80k or 80k α, tan ot ( 90 α ) otα ( 90 α ) tanα 90 + ; 80k, k ( k any integer). (4) For any angle egree measure α 60 k + 80 an α 80k + 90( k any integer), Page of 9

3 α tan tan α. (5) α tan A Lemma an its proof Lemma.. Suppose that a,,, are real numers suh that 0, a + 0, an + 0. ( In partiular, this hypothesis is satisfie when a,,, are positive.) Then, a a if, an only if, a + +. a Proof. Suppose that k. Then a k an k. We have ( k ) a k a + k + ( k + k k + (6) an also + ( ) k ( + ) k +. (7) a From (6) an (7), it follows that. a + + Conversely, assume that a a + +, whih implies ( a )( + ) ( a + )( ) a + a a a +, an thus, a; an sine a a 0, we otain ;. 4 Solving an SAS triangle using the Law of Sines Suppose that the lengths a an, as well as the egree measure, are given in Figure. From the Law of Sines () we otain Page of 9

4 a sinϕ (8) Sine θ an ϕ are egree measures of triangle angles, we have 0 < θ, ϕ < 80, an so sin θ > 0 an sin ϕ > 0; an, of ourse, a > 0, > 0. By Lemma, (8) implies sinϕ a, an y ientities (), + sinϕ a + sin sin θ ϕ θ + ϕ ( ) os ( ) a θ + ϕ θ ϕ ( ) os ( ) a + (9) Sine θ, ϕ, are triangle angle egree measures, the following hol true: θ + ϕ + 80 an θ + ϕ 0 < θ, ϕ, < 80. From whih we otain θ + ϕ 80 ; 90 an 0 < < 80 0 < < < < 0 0 < 90 < 90. This then shows that the onitions of ientity (5) are satisfie an therefore, θ + ϕ tan ot an θ + ϕ ot tan (0) ϕ Also, note that form 90 < < 0 an 0 < θ < 90 one easily otains θ ϕ 90 < < 90. () From (9) an (0) we infer that, θ ϕ a tan ot a + () A well-known fat from trigonometry is that for every real numer ν, there is a unique egree measure t suh that 90 < t < 90 an tan t ν. In other wors, t artanν or t tan ν with t measure in egrees. Let us rewrite 90 < t < 90 an tan t ν () Now, given a,, an, we first alulate the value ν of the right-han sie of (): Page 4 of 9

5 a ν ot (4) a + ν tan ( ) a a + Note that a alulator only has the tan key, whih means that we annot alulate the value of ot iretly. Now, let t e the unique egree measure suh that tan t ν an with 90 < t < 90. Then, y (), (), (), an (4) it follows that θ ϕ t ; θ ϕ t. an alsoθ + ϕ 80 (5) From (5) we otain the formulas, θ 90 + t, ϕ 90 t + (6) Rememer that after we ompute the value of ν in (4) y putting our alulator in egree moe, we then use the tan key to fin the value of t; an y the use of (6), we etermine the egree values of θ an ϕ. Finally, from a sin sin the Law of Sines () we an fin the value of the sielength y using either of or. sinφ 5 Examples. Solve the SAS triangle that satisfies tan, a, an. Solution: In this example, the egree measure is not given iretly. By (5), we have 4 4 tan. This tells us that sine is a triangle angle, 90 < < 80. By pressing that 4 Page 5 of 9

6 4 tan key on key on the alulator in the egree moe, we otain tan otain the (approximate) value of, we must a 80 to the answer otaine: Thus, to Next we ompute the value of ν in (4) with ot, a, an. We fin ν an so, 4 tan t Applying formulas (6) givesθ an ϕ Two eimal plae approximations for the aove angles are 6.87, θ 40.60, ϕ. 5 (with the sum eing exatly equal to 80). Using the alulator again, we fin that sin 0. 8 an Hene, ( )( 0.8) a sin What preise onition must the angle, an the lengths a an satisfy if t 45? Solution: If t 45, then ν tan t tan 45, an so y (4) we otain a tan. a + If we sustitute for tan a in (5) we otain, after some asi algera tan, a is the preise onition sought after. a. Suppose that r is a given numer greater than ; an that r.in other wors, we require that the sielengths a an have a presrie ratio r >. Also, let 60. (i) Fin the exat values of an sinϕ in terms of r. (ii) Express the thir sielength in terms of r an. (iii) Fin the exat values of θ an ϕ when r. (iv) What an e sai aout the angles θ an ϕ, if we allow r to vary in the open interval (,+ )? Page 6 of 9

7 Solution: (i) From r a an ot ot( 0) ( ) ( ) r we otain ν ; r + r Any y () we arrive at tan t > 0, sine r >. Sine 90 < t < 90 an r + tan t > 0, it follows that 0 < t < 90, whih implies sin t > 0 an os t > 0. We apply the ientity se t tan t + ; whih implies, sine os t > 0 an se t os, that t os t tan r + ( r ) + ( r + ) algera t + r r + + Next, t ( ost)( tan t) ( r ) ( ) ( ). r + r r + r sin ( ) ( ) + r + r + r + Now apply the well known ientity sin( 90 + ) os with t. By (6), we have os t os t os + sin t sin ; an y putting α os os (0),sin sin (0), an employing the expressions we otaine aove for sin t an ost in terms of r, we get (after some areful algera) α α r ( r ) + ( r + ) r. r r + From the Law of Sines () we also have sinϕ a r ; or equivalently, sinϕ (y the aove formula) r sinφ ( r ) + ( r + ) r. r + Page 7 of 9

8 (ii) From the Law of Sines, sin sin 60 sin ϕ sinϕ sinϕ ( r ) + ( r + ) r r +. (iii) When r,we otain from the aove formulas sin θ an sin ϕ ; θ 90 an ϕ 0. (iv) We invite the reaer to verify the following laims y using the fat that the quarati trinomial in r, f ( r) ( r ) + ( r + ) ; f ( r) 4r 4r + 4, r ; f r 4 r +., whih ours at ( ) Thus, on the open interval (,+ ), the trinomial f ( r) has asolute minimum value is inreasing in value. Aoringly, as r is ereasing in value towar, the angle measure ϕ ereases towar the value of 60 ; an sine θ 80 ϕ ϕ 0 ϕ, the egree measure θ is inreasingly approahing the value 60, so the limiting position of the triangle at han, is an equilateral one. On the other han, as r inreases in value towar, the angle measure ϕ ereases towar 0, while θ inreasingly approahes 90. Of ourse, when r, we have the exat values ϕ 0 an θ 90. As r inreases away from towar +, ϕ ereases towar 0, while θ inreases towar 0. Page 8 of 9

9 Referenes [] Roert Blitzer, Prealulus Essentials, n Eition, Prentie-Hall, In., 007, pp , 708 pp. Page 9 of 9