Implementing the Law of Sines to solve SAS triangles

June 8, 2009

Konstantine Zelator Dept. of Math and Computer Science1 Rhode Island College 600 Mount Pleasant Avenue Providence, RI 02908 U.S.A. e-mail : kzelator@ric.edu konstantine_zelator@yahoo.com

Introduction

Most trigonometry or precalculus texts offer a fairy brief treatment of the subject of solving a triangle; while only a few contain a more extensive analysis of the subject matter. By “solving a triangle” , we refer to the problem of determining all three sidelengths and interior triangle angle on given information on some of the triangle’s angles and/or sidelengths. Depending
1Effective August 1, 2009 and for the academic year 2009-2010: Konstantine Zelator Department of Mathematics 301 Thackeray Hall 139 University Place University of Pittsburgh Pittsburgh, PA 15260 e-mail: kzet159@pitt.edu

on the specific information, a unique triangle may be formed, more than one triangle, or no triangle at all. Robert Blitzer’s book, Precalculus, is among those books that contain a typical treatment of this subject (see [1]). Now, the case SAS refers to the situation wherein two sidelengths are given, as well as the degree measure of the angle contained between the two sides. As we know from Euclidean geometry, such a triangle is uniquely determined; meaning that any two triangles constructed with these specifications must be congruent. It appears that not only Blitzer’s book, but universally, in any book or text with the SAS case, the Law of Cosines is employed. For example, let us say that the lengths a and b are given; as well as the degree measure ω of the angle BCA ∠ (see Figure 1). Then, by using the Law of Cosines, one calculates the value of c , the length of BA ; and by using the
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Law of Sines, one determines the values of ϕ sin and θ sin . By the use of the inverse function on a calculator if necessary, one determines the degree measures ϕ and θ . Figure 1

In this work, we present an alternative approach to solving an SAS triangle. An approach whose key ingredient is the Law of Sines.

Law of Sines

     

= ω ϕ θ sin sin sin c b a

(1)

2 Basic trigonometric identities

The following identities are widely very well known and can be found in any standard trigonometry or precalculus text. For any angle degree measures α and β ,

( ) ( )       +

−

β α β α β α β α β α β α sin cos cos sin sin sin sin cos cos cos

(2) For any degree measures α and β ,

                  +       −

−       −       +

2 cos 2 sin 2 sin sin 2 cos 2 sin 2 sin sin β α β α β α β α β α β α

(3) For any angle degree measure α not of the form k 180 or k 180 90 + ; ( ) integer 180 90 , 180 any k k k + ≠ α ,

( ) ( )      

−

− α α α α tan 90 cot cot 90 tan .

(4) For any angle degree measure 180 360 + ≠ k α and ) integer ( 90 180 any k k + ≠ α , C B A a b c ω
ϕ
θ
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. 2 tan 1 2 tan 2 tan 2       −      

α α α

(5)

3
A Lemma and its proof

Lemma 3.1. Suppose that d c b a , , , are real numbers such that 0 , 0 ≠ + ≠ b a bd , and 0 ≠

• d c . ( In particular, this hypothesis is satisfied when d c b a , , , are positive.) Then, d c b a = if, and only if, d c d c b a b a

−

− .

Proof. Suppose that .k d c b a

= Then bk a = and dk c = . We have

( ) 1 1 1 ( 1 + −

−

−

− k k k b k b b bk b bk b a b a

(6) and also

( ) ( ) 1 1 1 1 + −

−

− k k d d d c d c d c d c .

(7) From (6) and (7) , it follows that . d c d c b a b a + −

−

Conversely, assume that d c d c b a b a + −

− , which implies
( )( ) ( )( ) , bd bc ad ac bd bc ad ac d c b a d c b a − + −

− − + ⇔ − +

− and thus, ; 2 2 ad bc = and since 0 ≠ bd , we obtain . ; 2 2 d c b a bd ad bd bc

=

4 Solving an SAS triangle using the Law of Sines Suppose that the lengths a and b , as well as the degree measure ω , are given in Figure 1. From the Law of Sines (1) we obtain Page 4 of 9

ϕ θ sin sin

b a

(8) Since θ and ϕ are degree measures of triangle angles, we have , 180 , 0 < < ϕ θ and so 0 sin

θ and ;0 sin

ϕ and, of course, .0 , 0

b a By Lemma 1, (8) implies

, sin sin sin sin b a b a + −

− ϕ θ ϕ θ and by identities (3),

( ) ( ) ( ) ( ) b a b a + −

− + + − 2 2 2 2 cos sin 2 cos sin 2 ϕ θ ϕ θ ϕ θ ϕ θ

(9) Since ω ϕ θ , , are triangle angle d

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