Download ANALYTICAL AND NUMERICAL SOLVING OF GENERAL

Poland
Problems of Applied Sciences, 2015, Vol. 3, pp. 015 – 032
Szczecin

dr n. tech. Andrzej Antoni CZAJKOWSKIa,b, Kamil SIDORKIEWICZa
a
University of Szczecin, Faculty of Mathematics and Physics, Department of Informatics and Technical Education (graduate)
Uniwersytet Szczeciński, Wydział Matematyczno-Fizyczny, Katedra Edukacji Informatycznej i Technicznej (absolwent)
b
Higher School of Technology and Economics in Szczecin, Informatics and Technical Education
Wyższa Szkoła Techniczno-Ekonomiczna w Szczecinie, Edukacja Informatyczno-Techniczna
ANALYTICAL AND NUMERICAL SOLVING OF GENERAL
TRIANGLES WITH APPLICATION OF NUMERICAL PROGRAMS
MS-EXCEL, MATHCAD AND MATHEMATICA
Abstract
Introduction and aims: The paper shows the analytical models of solving some general triangles
with appropriate discussion. For general triangles have been discussed four cases. The main aim
of this paper is not only to create some analytical algorithms for solving general triangle, but also
their implementation in programs MS-Excel, MathCAD and Mathematica.
Material and methods: Elaboration of four analytical cases of solving the general triangles has
been made on the basis of the relevant trigonometric properties occurring in a general triangle. In
the paper have been used some analytical and numerical methods by using MS-Excel, MathCAD
and Mathematica programs.
Results: As some results have been obtained numerical algorithms in the programs MS-Excel,
MathCAD and Mathematica for four analytical cases of solving the general triangles.
Conclusion: Created numerical algorithms of solving the general triangles in the programs MSExcel, MathCAD and Mathematica allow for faster significant performance calculations than the
traditional way of using logarithms and logarithmic tables.
Key-words: Trigonometry, solving of general triangles, numerical algorithms, MS-Excel,
MathCAD, Mathematica.
(Received: 01.05.2015; Revised: 15.05.2015; Accepted: 01.06.2015)
ANALITYCZNO-NUMERYCZNE ROZWIĄZYWANIE TRÓJKĄTÓW
DOWOLNYCH Z ZASTOSOWANIEM PROGRAMÓW
MS-EXCEL, MATHCAD i MATHEMATICA
Streszczenie
Wstęp i cele: W pracy pokazano analityczne modele rozwiązywania trójkątów dowolnych wraz z
odpowiednią dyskusją. Dla trójkątów dowolnych omówiono cztery przypadki. Głównym celem jest
pracy jest nie tylko utworzenie algorytmów analitycznych rozwiązywania takich trójkątów lecz
również ich implementacja w programach MS-Excel, MathCAD i Mathematica.
Materiał i metody: Opracowanie czterech analitycznych przypadków rozwiązywania trójkątów
dowolnych wykonano opierając się odpowiednich własnościach trygonometrycznych występujących w trójkącie dowolnym. Zastosowano metodę analityczną i numeryczną wykorzystując programy MS-Excel, MathCAD i Mathematica.
Wyniki: Otrzymano algorytmy numeryczne w programach MS-Excel, MathCAD i Mathematica
dla czterech analitycznych przypadków rozwiązywania trójkątów dowolnych.
Wniosek: Utworzone algorytmy numeryczne rozwiązywania trójkątów dowolnych w programach
MS-Excel, MathCAD oraz Mathematica, pozwalają na znaczne szybsze wykonanie obliczeń niż
drogą tradycyjną z użyciem logarytmów i tablic logarytmicznych.
Słowa kluczowe: Trygonometria, rozwiązywanie trójkątów dowolnych, algorytmy numeryczne,
MS-Excel, MathCAD, Mathematica
(Otrzymano: 01.05.2015; Zrecenzowano: 15.05.2015; Zaakceptowano: 01.06.2015)
© A.A. Czajkowski, K. Sidorkiewicz 2015
Trigonometry / Trygonometria
A.A. Czajkowski, K. Sidorkiewicz

1. Introduction to trigonometry
1.1. Types of triangles and basic properties
Before we begin the analysis of solving the general triangles, let us look at the types triangles (Fig. 1) and the basic conditions of existence of the triangle (Theorem 1).
Triangles division of due to
angles
sides
two
sides
equal
all sides
of
different
lengths
equilateral
isosceles
triangle
triangle
differentsided
all
interior
angles
acute
one
right
angle
one
obtuse
angle
all
sides
equal
acute
triangle
rightangled
triangle
obtuse
triangle
triangle
a
b
c
a
b
•
a
c
c
b
a
a
a
a
b
b
c
b
a
Fig. 1. Types of triangles
Source: Elaboration of the Authors basing on [4], [8], [20], [22] & [28]
Theorem 1. (Basic properties in the triangle) [2], [6]
In any length of each side of the triangle is less than the sum of the length of the other side
and higher than the absolute value of the difference in length of the sides.
This fact is written by the following conditions:
16
|b − c| < a < b + c ,
(1)
|a −c| < b < a + c ,
(2)
|a − b| < c < a + b .
(3)
Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

1.2. Properties in general triangle
The general triangle has six basic elements, i.e. three sides a, b and c and three angles α, β
and γ. The solution of general triangle involves the determination of three basic elements form
six elements, if we have three other. To solve the general triangles should be used theorems of
sines, cosines and tangents of half angle function.
• Theorem of sines and cosines
In solving the general triangles are used of sines and cosines formulas. These formulas are
presented in the sines and cosines theorems.
Theorem 2. Theorem of sines (Snellius)
The sides and angles of any triangle satisfy the following system of equations [5], [10]-[14]:
α + β + γ = 180°,
(4)
a
b
c
=
=
.
sin(α) sin(β) sin( γ )
(5)
Theorem 3. Theorem of sines (Snellius)
In any triangle, the ratio of the length of any side of the triangle to the sine of the angle
opposite this side is constant and equal to the diameter of the circle which is circumscribed on
this triangle [5], [10]-[14]:
a
b
c
=
=
= 2R ,
sin(α) sin(β) sin( γ )
(6)
where R is the radius of the circle circumscribed on a given triangle (Fig. 1). The equality (6)
is called the formula of sines.
C
γ
c
b
α
O
2R
β
B
a
A
Fig. 2. A circle circumscribed on the triangle
Source: Elaboration of the Authors basing on [10]
17
A.A. Czajkowski, K. Sidorkiewicz

Theorem 4. Theorem of cosines (Carnot)
In any triangle, the square of the length of any side equals the sum of the squares of the
lengths of the other two sides decreased by twice the product of these two sides lengths and
the cosine of their included angle [4], [17], [24], [25].
Theorem 5. Theorem of cosines (Carnot)
The sides and angles of any triangle satisfy the following system of equations [4], [24], [25]:
a 2 = b 2 + c 2 − 2bc ⋅ cos(α),
(7)
b 2 = a 2 + c 2 − 2ac ⋅ cos(β),
(8)
c 2 = a 2 + b 2 − 2ab ⋅ cos( γ ).
(9)
The equalities (7) - (9) are called the formulas of cosines.
Corollary 1.
If in the formulas of cosines one of the angles is a right angle, the cosine theorem takes on
its familiar form known as the Pythagorean theorem [4], [17], [24], [25].
a 2 = b2 + c2 ,
(10)
b 2 = a 2 + c2 ,
(11)
c2 = a 2 + b 2 .
(12)
• Theorems on any triangle
Theorem 6. (Regiomontan)
The sides and angles of any triangle satisfy the following system of equations [4], [8], [20]:
α −β
tg
=
 2 
α−γ
tg
=
 2 
a−b γ
ctg  ,
a + b 2
(13)
a − c β
ctg  ,
a + c 2
(14)
β − γ b −c α
tg
ctg  .
=
 2  b+c 2
(15)
The equalities (13) - (15) are called the formulas of tangents.
Theorem 7. (The half-angle)
The sides and angles of any triangle satisfy the following system of equations [4], [8], [20]:
α
tg  =
2
(p − b)(p − c)
p(p − a)
(16)
(p − a)(p − c)
β
tg  =
p(p − b)
2
(17)
γ
tg  =
2
(18)
(p − a)(p − b)
p(p − c)
Equalities (16) - (18) are called the formulas of the half angle.
18
Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

1.3. Cases of solving the general triangles
For the general triangle ABC shown in figure 2, there are four cases of solving which are
given below (Tab. 1). Let us consider any given general triangle ABC.
C
γ
b
A
a
β
α
B
c
Fig. 3. Scheme of the triangle ABC
Source: Elaboration of the Authors
For the general triangle ABC, shown in figure 3, there are four cases of solving which are
given below in the table 1.
Table 1. Cases of solving the general triangles [4], [8], [20]
Case:
Data:
Unknown:
a, β, γ
b, c, α
side, angle, angle
side, angle, angle
Side and angles adjacent to it
Two sides and angle
a, b, α
c, β, γ
side, side, angle
side, angle, angle
The two sides and the contained angle between them
Side and two angles
a, b, γ
c, α, β
side, side, angle
side, angle, angle
The two sides and the angle adjacent to one of them
Side and two angles
a, b, c
α, β, γ
side, side, side
angle, angle, angle
Three sides
Three angles
I
II
III
IV
Source: Elaborated by the Authors
1.4. Discussion of general triangles solutions
In the general triangle there is six characterized sizes of this triangle (three angles α, β, γ
and three sides a, b, c). If we have three given elements of a general triangle, the other three
can be determined in accordance with the table 2.
19
A.A. Czajkowski, K. Sidorkiewicz

Table 2. Discussion of solving the general triangles [4], [5], [8], [19], [20], [22], [28]
Case:
Given data:
I
a, β, γ
Solution:
α = 180° − (β + γ) , b = a ⋅ sin(β) ,
c=
sin(α)
a ⋅ sin( γ )
sin(α)
One solution
γ = 180° − (α + β)
1) a > b
c=
II
a, b, α
sin(β) =
Two solutions
β1
γ1 = 180° − (α + β1)
b ⋅ sin( α )
a
Hence we calculate
a ⋅ sin( γ )
sin(α)
2) b⋅sin(α) < a < b
c1 =
a ⋅ sin( γ1 )
sin(α)
β2 = 180° − β1
γ2 = 180° − (α + β2)
the angle β
c2 =
a ⋅ sin( γ 2 )
sin(α)
3) a = b⋅sin(α)
One solution
β = 90°
γ = 90° − α
c = b⋅cos(α)
4) a < b⋅sin(α)
There is no any solution
1) c2 = a2 + b2 – 2ab⋅cos(γ) Hence we calculate a length of the side c
If a > b, then sin(β) =
III
a, b, γ
b ⋅ sin( γ )
c
Thence we calculate the angle β
α = 180° − (β + γ)
2) tg α − β  = a − b ctg γ 
 2  a + b 2
Hence we calculate
having α + β = 90° − γ we find the angles: α, β
2
2
α −β
and
2
Side c =
a ⋅ sin( γ )
sin(α)
2
2
2
b 2 + c2 − a 2
a 2 + b 2 − c2
, cos(β) = a + c − b , cos(γ ) =
2bc
2ab
2ac
Hence we calculate the angles: α, β, γ
1) cos(α) =
IV
a, b, c
2) a + b + c = 2p (perimeter of the triangle)
α
tg  =
2
(p − b)(p − c) ,  β 
(p − a)(p − c) ,
tg  =
p(p − a)
p(p − b)
2
γ
tg  =
 2
(p − a)(p − b) Thence we calculate the angles: α, β, γ
p(p − c)
Source: Elaborated by the Authors
20
Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

2. Solving of general triangles
2.1. The first case – Given one side and two angles adjacent to that side
Solve any triangle having the side and angles adjacent to this side (in short ASA).
2.1.1. Theoretical analysis
C
a
γ
b
β
α
A
B
c
Fig. 4. Scheme of general triangle ABC with given side c and the angles α and β
Source: Elaboration of the Authors
Data in ∆ ABC: The length of the side c and measure of the angles α and β.
Search in ∆ ABC: The length of the sides a and b and measure of the angle γ.
Solution:
♦ Based on the properties of the angles in given triangle we calculate the angle γ:
γ = π − ( α + β) .
♦ Using the theorem of sines we determine the side a:
a
c
=
sin(α) sin( γ )
hence
c ⋅ sin(α)
a=
sin( γ )
(19)
(20)
(21)
and after taking into account the formula (19) we obtain:
a=
c ⋅ sin(α)
.
sin[π − (α + β)]
(22)
♦ Using again the theorem of sines we determine the side b:
b
c
=
sin(β) sin( γ )
thence
b=
(23)
c ⋅ sin(β)
sin(γ )
(24)
and after taking into account the formula (19) we have:
c ⋅ sin(β)
.
sin[π − (α + β)]
Answer: The requested unknowns are finally defined by the following formulas:
b=
sides
a=
c ⋅ sin(α)
,
sin[π − (α + β)]
b=
c ⋅ sin(β)
sin[π − (α + β)]
and angle
(25)
γ = π − ( α + β) .
21
A.A. Czajkowski, K. Sidorkiewicz

2.1.2. Numerical algorithms in MS-Excel, MathCAD and Mathematica programs
For numerical analysis, we assume that in the general triangle we have the following data:
length of the side c = 13,5 and measure of the angles α = 44° and β = 38,641°.
♦ Program MS-Excel [3], [9], [16], [18], [23]
Algorithm:
Commentary:
B2=13,5
Given length of the side c
B3=44
Given measure of the angle α [º]
B4=38,641
Given measure of the angle β [º]
B5=(B3*PI())/180
Command for the angle α=0,767945[rad]
B6=(B4*PI()/180)
Command for the angle β=0,674413[rad]
B7=PI()-(B5+C6)
Command for the angle γ=1,699 [rad]
B8=B7*180/PI()
Command for the angle γ=97,359 [°]
B9=B2*SIN(B5)/SIN(B7)
Command for a side a
B10=B2*SIN(B6)/SIN(B7)
Command for a side b
97,359
Result: calculated measure of the angle γ [°]
9,455775
Result: calculated length of the side b
8,499936
Result: calculated length of the side a
♦ Program MathCAD [15], [21], [29]
Algorithm:
Commentary:
c:=13.5
Given length of the side c
α0:=44
Given measure of the angle α [º]
β0:=38.641
Given measure of the angle β [º]
α :=
α0 ⋅ π
= 0.767945
180
Command and result: calculated measure of the angle α [rad]
β :=
β0 ⋅ π
= 0.674413
180
Command and result: calculated measure of the angle angle β [rad]
γ := π − (α + β) = 1.699
Command and result: calculated measure of the angle angle γ [rad]
γ :=
α ⋅ 180
= 97.359
π
Command and result: calculated measure of the angle angle γ [°]
a :=
c ⋅ sin(α)
= 9.455775
sin( γ )
Command and result: calculated length of the side a
b :=
c ⋅ sin(β)
= 8.499936
sin( γ )
Command and result: calculated length of the side b
22
Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

♦ Program Mathematica [1], [7], [26], [27]
Algorithm:
Commentary:
c:=13.5
Given length of the side c
A0:=44
Given measure of the angle α [º]
B0:=38.641
Given measure of the angle β [º]
A=N[A0*Pi/180];
Command for the angle α=0,767945[rad]
B=N[B0*Pi/180];
Command for the angle β=0,674413[rad]
G=Pi-(A+B);
Command for the angle γ=1,69924 [rad]
G1=A*180/Pi
Command for the angle γ [°]
a=c*SIN(A)/SIN(G)
Command for the side a
b=c*SIN(B)/SIN(G)
Command for the side b
97.359
Result: calculated measure of the angle γ [°]
9.45577
Result: calculated length of the side a
8.499936
Result: calculated length of the side b
♦ Summary of the results
Data:
Result of calculation:
side c
angle α [º]
angle β [º]
angle γ [º]
side a
side b
13,5
44
38,641
97,359
9,45577
8,499936
Verification:
α + β + γ = 180º
⇔ 44º + 38,641º + 97,359º = 180º
|b − c| < a < b + c
⇔
5,00006 < 9,45577 < 21,999936
|a − c| < b < a + c
⇔
4,04423 < 8,499936 < 22,95577
|a − b| < c < a + b
⇔
0,955834 < 13,5 < 17,955706
2.2. The second case – Given two sides and the angle included between them
Solve the general triangle having two sides and the angle included between them (in short SAS).
2.2.1. Theoretical analysis
C
b
a
γ
β
α
A
B
c
Fig. 5. Scheme of the triangle ABC with given sides a and c and angle γ
Source: Elaboration of the Authors
23
A.A. Czajkowski, K. Sidorkiewicz

Data in ∆ABC:
The length of the sides a and b and measure of the angle γ.
Search in ∆ABC: The length of the side c and measure of the angles α and β.
Solution:
♦ Using the theorem of cosines we determine the side c:
c2 = a 2 + b 2 − 2ab ⋅ cos( γ ) ,
(26)
c = a 2 + b 2 − 2ab ⋅ cos( γ ) .
(27)
hence
♦ Using the theorem of sines we determine the angle α:
c
a
=
,
sin( γ ) sin(α)
(28)
a
sin( γ ) .
c
(29)
hence
sin(α) =
Using the inverse sine function, we have:
a

α = arcsin  sin( γ ) .
c

(30)
After taking the formula (27) in (30) we get:

a ⋅ sin( γ )
α = arcsin 
 a 2 + b 2 − 2ab ⋅ cos( γ )


.


(31)
♦ From property angles of a triangle we have:
β = π − ( α + γ ),
(32)
thus


asin( γ )

β = π − α + arcsin 

 a 2 + b 2 − 2ab ⋅ cos( γ )




.

(33)
Answer: The requested unknowns are finally defined by the following formulas:
side
angels
c = a 2 + b 2 − 2ab ⋅ cos( γ )

a ⋅ sin( γ )
α = arcsin 
 a 2 + b 2 − 2ab ⋅ cos( γ )




a ⋅ sin( γ )
 , β = π −  γ + arcsin



 a 2 + b 2 − 2ab ⋅ cos( γ )



2.2.2. Numerical algorithms in programs MS-Excel, MathCAD and Mathematica
For numerical analysis, we assume that in the general triangle we have the following data:
length of the sides a = 9,455775 and b = 8,499936 and measure of the angle γ = 97,359°.
Note: If a > b, then there is one solution (Tab. 2).
24

.


Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

♦ Program MS-Excel [3], [9], [16], [18], [23]
Algorithm:
Commentary:
B2=9,455775
Given length of the side a
B3=8,499936
Given length of the side b
B4=97,359
Given measure of the angle γ [º]
B5=97,359*PI()/180
Command for the angle γ=1,699 [rad]
B6=PIERWIASTEK(B2^2+B3^2-2*B2*B3*COS(B5))
Command for the side c
B7=ASIN(B2*SIN(B5)/B6)
Command for the angle α [rad]
B8=PI()-(B5+B7)
Command for the angle β [rad]
B9=B7*180/PI()
Command for the angle α [°]
B10=B8*180/PI()
Command for the angle β [°]
13,5
Result: calculated measure of the side c
44
Result:calculated measure of the angle α [°]
38,641
Result: calculated measure of the angle β [°]
♦ Program MathCAD [15], [21], [29]
Algorithm:
a:=9.455775
b:=8.499936
γ1:=97.359
γ1 ⋅ π
= 1.699235
γ :=
180
Commentary:
Given length of the side a
Given length of the side b
Given measure of the angle γ [º]
Command and result: calculated measure
of the angle γ [rad]
c := (a 2 + b 2 − 2abcos(γ ))0.5 = 13.500001
Command and result: side c
 a ⋅ sin( γ ) 
α1 := asin
 = 0.767
c


Command and result: calculated measure
of the angle angle α [rad]
β1 := π − (α + γ ) = 0.674413
Command and result: calculated measure
of the angle angle β [rad]
Command and result: calculated measure
of the angle angle α [°]
α1 ⋅ π
= 44
180
β1 ⋅ π
β :=
= 38.641
180
α :=
Command and result: calculated measure
of the angle angle β [°]
♦ Summary of the results
Data:
side a
9,455775
side b
8.499936
Result of calculation:
angle γ [º]
97,359
side c
13,5
angle α [º]
44
angle β [º]
38,641
Verification:
α + β + γ =180º
|b − c| < a < b + c
|a − c| < b < a + c
|a − b|< c < a +b
⇔
⇔
⇔
⇔
44º + 38,641º + 97,359º = 180º
5, 000064 < 9, 455775 < 21,999936
4, 044225 < 8, 499936 < 22,955775
0,955839 < 13,5 < 17,955711
25
A.A. Czajkowski, K. Sidorkiewicz

♦ Program Mathematica [1], [7], [26], [27]
Algorithm:
Commentary:
a:=9.455775
Given length of the side a
b:=8.499936
Given length of the side b
G0=97.359
Given measure of the angle γ [º]
G:=G0*Pi/180
Command for the angle γ=1,699 [rad]
c=Sqrt[a^2+b^2-2*a*b*Cos[G]]
Command for the side c
A=ArcSin[a*Sin[G]/c]
Command for the angle α [rad]
B=Pi-(G+A)
Command for the angle β [rad]
A*180/Pi;
Command for the angle α [°]
B*180/Pi;
Command for the angle β [°]
13.5
Result: calculated length of the side c
44
Result: calculated measure of the angle α [°]
38.641
Result: calculated measure of the angle β [°]
2.3. The third case – Given two sides and angle adjacent to one of these sides
Solve the general triangle having two sides and angle adjacent to one of these sides (in short ASS).
2.3.1. Theoretical analysis
C
b
γ
a
β
A
α
B
c
Fig. 6. Scheme of the triangle ABC with given sides a and b and angle α
Source: Elaboration of the Authors
Data in ∆ABC:
The length of the sides a and b and measure of the angle α.
Search in ∆ABC: The length of the side c and measure of the angles β and γ.
Solution:
♦ Using the theorem of sines we determinate the angle β :
a
b
=
,
sin(α) sin(β)
(34)
Hence using the inverse sine function, we have:
 b ⋅ sin(α) 
β = arcsin 
.
a


26
(35)
Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

♦ From the property angles of a triangle we have
γ = π − ( α + β),
(36)

 b ⋅ sin(α)  
γ = π − α + arcsin 
 .
a



(37)
hence
♦ Using the theorem of cosines we determine the side c:
c 2 = a 2 + b 2 − 2ab ⋅ cos( γ ) .
(38)
♦ After taking the formula (37) in (38) we obtain:
 
 b ⋅ sinα   
c 2 = a 2 + b 2 − 2ab ⋅ cos  π −  α + arcsin 
 
 a  
 
which, after appropriate transformations gives:

 b ⋅ sin(α)  
c = a 2 + b 2 + 2ab ⋅ cosα + arcsin 
 .
a



(39)
(40)
Answer: The requested unknowns are finally defined by the following formulas:
side &
angels


 b ⋅ sin(α)  
 b ⋅ sin(α )  
 b ⋅ sin(α) 
c = a 2 + b 2 + 2ab ⋅ cos α + arcsin 
, γ = π − α + arcsin 
 , β = arcsin
 .


a
a
a








2.3.2. Numerical algorithms in programs MX-Excel, MathCAD and Mathematica
For numerical analysis, we assume that in the general triangle we have the following data:
length of the sides a = 9,455775 and b = 8,499936 and measure of the angle α = 44°
♦ Program MS-Excel [3], [9], [16], [18], [23]
Algorithm:
Commentary:
B2=9,455775
Given length of the side a
B3=8,499936
Given length of the side b
B4=44
Given measure of the angle α [º]
B5=B4*PI()/180
Command for the angle α [rad]
B6=ASIN(B3+SIN(B5)/B2)
Command for the angle β [rad]
B7=B6*180/PI()
Command for the angle β [º]
B8=PI()-(B5+B6)
Command for the angle γ [rad]
B9=B8*PI()/180
Command for the angle γ [°]
B10=PIERWIASTEK(B2^2+B3^2-2*B2*B3*COS(B5))
Command for the side c
38,641
Result: calculated measure of the angle β [°]
97,359
Result: calculated measure of the angle γ [°]
13,5
Result: calculated length of the side c
27
A.A. Czajkowski, K. Sidorkiewicz

♦ Program MathCAD [15], [21], [29]
Algorithm:
Commentary:
a: = 9.455775
Given length of the side a
b: = 8.499936
Given length of the side b
α: = 44
Given measure of the angle α [º]
α1 :=
α⋅π
= 0.767945
180
Command and result: calculated measure of the angle α [rad]
 b ⋅ sin(α1) 
β1 := asin 
 = 0.674413
a


β :=
β1 ⋅ 180
= 38.641
π
Command and result: calculated measure of the angle β [°]
γ1 := π − (α1 + β1) = 1.699235
γ :=
Command and result: calculated measure of the angle β [rad]
Command and result: calculated measure of the angle γ [rad]
γ1 ⋅ 180
= 97.359
π
Command and result: calculated measure of the angle γ [°]
c := a 2 + b 2 − 2 ⋅ a ⋅ b ⋅ cos( γ1) = 13.5
Command and result: calculated length of the side c
♦ Program Mathematica [1], [7], [26], [27]
Algorithm:
a:=9.455775
Given length of the side a
b:=8.499936
Given length of the side b
A=44
Given measure of the angle α [º]
A1:=A*Pi/180
Command for the angle α [rad]
B1:=ArcSin[(b*Sin[A1]/a]
Command for the angle β [rad]
B=B1*180/Pi
Command for the angle β [°]
G1=Pi-(A1+B1);
Command for the angle γ [rad]
G=G1*180/Pi
Command for the angle γ [°]
c=Sqrt[a^2+b^2-2*a*b*Cos[G]]
Command for the side c
38,641
Result: calculated measure of the angle β [°]
97,359
Result: calculated measure of the angle γ [°]
13,5
Result: calculated length of the side c
Commentary:
♦ Summary of the results
Data:
side a
9,455775
side b
8.499936
Result of calculation:
angle α [º]
44
side c
13,5
angle β [º]
38,641
Verification:
α + β + γ = 180º ⇔ 44º + 38,641º + 97,359º = 180º
| b − c | < a < b + c ⇔ 5,00006 < 9,45577 < 21,999936
| a − c | < b < a + c ⇔ 4,04423 < 8,499936 < 22,95577
| a − b | < c < a + b ⇔ 0,955834 < 13,5 < 17,955706
28
angle γ [º]
97,359
Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

2.4. The fourth case – Given three sides
Solve the general triangle having three sides (in short SSS).
2.4.1. Theoretical analysis
C
b
A
a
γ
β
α
B
c
Fig. 7. Scheme of the triangle ABC with given sides a, b and c
Source: Elaboration of the Authors
Data in ∆ABC: The length of the sides a, b and c.
Search in ∆ABC: The measure of the angels α, β and γ.
Solution:
♦ Using the theorem of cosines we determinate the angle α:
a 2 = b 2 + c 2 − 2bc ⋅ cos(α) ,
(41)
hence
b 2 + c2 − a 2
cos(α) =
.
2bc
Using the inverse cosine function, we get:
 b2 + c2 − a 2 
.
α = arccos


2bc


(42)
(43)
♦ Using again the theorem of cosines we determinate the angle β:
b 2 = a 2 + c 2 − 2ac ⋅ cos(β) ,
(44)
hence
a 2 + c2 − b 2
.
2ac
Using again the inverse cosine function, we get:
cos(β) =
 a 2 + c2 − b2 
.
β = arccos


2ac


(45)
(46)
Taking into account a property angles of a triangle we determine the angle γ:
γ = π − (α + β)
(47)
Answer: The requested unknowns are finally defined by the following formulas:
Angels
2
2 
2

 b 2 + c2 − a 2 
 2 2
 2 2
 2 2
 , β = arccos a + c − b  , γ = π − arccos b + c − a  + arccos a + c − b  .
α = arccos








2bc
2ac
2bc
2ac









29
A.A. Czajkowski, K. Sidorkiewicz

2.4.2. Numerical algorithms in programs MX-Excel, MathCAD and Mathematica
For numerical analysis, we assume that in the general triangle we have the following data:
lengths of the sides: a = 9,455775; b = 8,499936 and c = 13,5.
♦ Program MS-Excel [3], [9], [16], [18], [23]
Algorithm:
Commentary:
B2=9,455775
Given length of the side a
B3=8,499936
Given length of the side b
B4=13,5
Given length of the side c
B5=ACOS(((B3^2+B4^2-B2^2)/(2*B3*B4))
Command for the angle α [rad]
B6=ACOS(((B2^2+B4^2-B3^2)/(2*B2*B4))
Command for the angle β [rad]
B7=PI()-(B5+B6)
Command for the angle γ [rad]
B8=B5*180/PI()
Command for the angle α [°]
B9=B6*180/PI()
Command for the angle β [°]
B10=B7*180/PI()
Command for the angle γ [°]
44,000001
Result: calculated measure of the angle α [°]
38,641003
Result: calculated measure of the angle β [°]
97,358993
Result: calculated measure of the angle γ [°]
♦ Program MathCAD [15], [21], [29]
Algorithm:
Commentary:
a:=9.455775
Given length of the side a
b:=8.499936
Given length of the side b
c:=13.5
Given length of the side c
 b2 + c2 − a 2 
 = 0.767945
α1 := acos
 2⋅b⋅c 


Command and result: calculated measure of the angle α [rad]
 a 2 + c2 − b2 
 = 0.674413
β1 := acos


2ac


Command and result: calculated measure of the angle β [rad]
γ1 := π − (α + β) = 1.699235
Command and result: calculated measure of the angle γ [rad]
α :=
α1 ⋅ 180
= 44.000004
π
Command and result: calculated measure of the angle α [°]
β :=
β1 ⋅ 180
= 38.641003
π
Command and result: calculated measure of the angle β [°]
γ :=
γ1 ⋅ 180
= 97.358993
π
Command and result: calculated measure of the angle γ [°]
30
Analytical and numerical solving of general triangles with application of numerical programs MS-Excel, MathCAD and Mathematica

♦ Program Mathematica [1], [7], [26], [27]
Algorithm:
Commentary:
a:=9.455775
Given length of the side a
b:=8.499936
Given length of the side b
c:=13.5
Given length of the side c
A1=ArcCos[(b^2+c^2-a^2)/(2*b*c)];
Command for the angle α [rad]
B1=ArcCos[(a^2+c^2-b^2)/(2*a*c)];
Command for the angle β [rad]
C1=Pi-(A1+B1);
Command for the angle γ [rad]
A1*180/Pi
Command for the angle α [°]
B1*180/Pi
Command for the angle β [°]
C1*180/Pi
Command for the angle γ [°]
44
Result: calculated measure of the angle α [°]
38.641
Result: calculated measure of the angle β [°]
97.359
Result: calculated measure of the angle γ [°]
♦ Summary of the results
Data:
side a
9,455775
Result of calculation:
side c
angle α [º]
angle β [º]
13,5
44
38,641
Verification:
α + β + γ = 180º ⇔ 44º + 38,641º + 97,359º = 180º
| b − c | < a < b + c ⇔ 5,00006 < 9,45577 < 21,999936
side b
8,499936
|a − c| < b < a + c
⇔
4,04423 < 8,499936 < 22,95577
|a − b| < c < a + b
⇔
0,955834 < 13,5 < 17,955706
angle γ [º]
97,359
3. Conclusions
• Discussed analytical models of solving the general triangles allow to perform the other
theoretical considerations in each of the four basic cases, taking into account the definition
of trigonometric functions and the corresponding properties and patterns.
• Created numerical algorithms of solving the general triangles in MS-Excel, MathCAD, and
Mathematica programs allow for significant execution of faster calculations than the traditional way of using logarithms and logarithmic tables.
• Problem of solving the general triangles finds wide applications in technical sciences and
related fields.
31
A.A. Czajkowski, K. Sidorkiewicz

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[5]
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[11]
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[15]
[16]
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[18]
[19]
[20]
[21]
[22]
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[26]
[27]
[28]
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32
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