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APPLICATION OF ELEMENTS OF
TRIGONOMETRY IN SOLUTION OF TRIANGLES
Otajonova Sitorabonu Shuxratovna
Asia International university
General technician sciences department intern teacher
sitorabonu_shukhratovna@mail.ru
ANNOTATION
This in the article of geometry school in the chair important role who plays
trigonometry department triangles to solve circle applications viewed Triangles in
solving three element giving , his the rest elements to find issues seen Such issues
in solving in trigonometry equivalent ratios support unknown elements to determine
method such as practical issues solve viewed
Key words :
Sinuses theorem , Cosines theorem , Molweide formula , triangles
in solving three in case issues .
THE USE OF TRIGONOMETRY ELEMENTS IN SOLVING
TRIANGLES
Otazhonova Sitorabonu Shukhratovna
Asian International University
Trainee teacher of the Department of General Technical Sciences
sitorabonu_shukhratovna@mail.ru
ABSTRACT
This paper discusses the implementation of the trigonometry section, which
plays an important role in the school geometry course concerning the solution of
triangles. When solving triangles, problems are constructed to find the rest of its
elements, giving three of its elements. When solving triangles, problems are
constructed to find the rest of its elements, giving three of its elements. When solving
such problems, it is possible to solve practical problems, such as the method of
determining equivalent ratios and unknown elements in trigonometry.
Keywords:
The sine theorem, the cosine theorem, the Mollweide formula,
three-case problems in solving triangles.
INTRODUCTION
of trigonometry appear to be in practice calculations , exactly different
geometric forms elements in finding given elements enough quantity according to
this elements to determine necessity the need with depend In ancient times Ancient
in Greece one row astronomical issues solve with depends calculations during
trigonometry field development important contribution added of trigonometry in the
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X- IX and XIII centuries Central Asia and Caucasian regions of scientists scientific
works and created works main important have
Science next development era that's it showed that trigonometric functions only
work in release not but count in geometry solve for necessary which hardware
functions ; as well as this functions mechanics and in physics periodic processes also
important in learning role plays So so , trigonometric functions to the theory based
on without analytical geometry direction appear it has been . Trigonometric of
functions geometric theory trigonometry practical to issues application to do more
suitable will come
MAIN PART
of the problem from the content come until it comes out triangles in solving
geometric apparently except of the matter to the classification have to be no doubt
These are as follows to the circumstances is divided into :
first case . Of the triangle two corner and one given a linear element let it be ;
Second case Of the triangle one corner and two given a linear element let it be
;
Third case Of the triangle three given a linear element let it be
Triangles to solve circle issues solve method according to the first in the case
issues directly row equivalent proportions tools with will be solved . Second in case
issues trigonometric to Eqs to the system is brought . This in the case in matters of
the triangle again one second corner to find need will be Otherwise in other words
relationship will be done . Third in case in matters of the triangle two
corner is found . In general so to speak , on condition the given element is a triangle
corner element if not , that is triangle sides when given, the problem becomes simpler
.
Issue 1.
ABC
in a triangle
(
)
b
c
78,
a
65,
R
28
r
is equal to if , then
b
and
c
find the
This is the third issue in case is an issue . Sinuses from the theorem
78
3
sin
130
5
3
2
5
a
R
relationships we can From this
4
cos
5
0
90
and
1
,
2
3
tg
but
2
2
P
r
a
tg
from
84,
2
P
a
162,
P
324
P
from that
246
b
c
is taken . Secondly
2
Pr
15120.
sin
sin
S
bc
So,
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246,
15120.
b
c
bc
b
c
that it was for
120,
b
126
c
come comes out
If
0
90
if so , then equations system for real didn't happen solution come
comes out
Reminder .
3
sin
5
and
4
cos
5
from being
3
2
tg
is equal to .
28
2
3
P
a
and
4
cos
5
from being
28
234
262
.
2
3
3
P
From this
524
3
P
a
b
c
and
524
290
78
.
3
3
b
c
Pr
524 28 5
73360
;
sin
3 3
9
bc
2
290
73360
0,
3
9
z
z
1,2
1
2
145
52335
,
,
; ,
.
3
z
z
b z
c b c
Issue 2.
In the triangle
2
2
2
4 3
R
S
b
c
relationship it is appropriate . Of
the triangle
a
side opposite
find the angle .
This issue is also third in case is an issue . 2
sin
S
bc
and
2
2
2
2
cos
b
c
a
bc
known in formulas
S
and
2
2
b
c
of value given to the
relationship if you put it , then
2
2
2 3
sin
2
cos
R
bc
a
bc
is taken . Here
, ,
a b c
line elements corner to the element if we replace ( i.e
2 sin ,
a
R
2 sin ,
a
R
2 sin
c
R
), then :
2
2
2
2
8 3
sin
sin
sin
4
sin
8sin
sin cos
R
R
R
or
2
R
abbreviated to :
2
1 8 3 sin
sin
sin
4sin
8sin
sin cos .
From this
2
1 4sin
8sin sin
cos
3 sin
;
2
1
1
3
sin
4sin
sin
cos
sin
4
2
2
or
0
0
1
1
sin
sin
4sin
sin
sin 30 cos
cos30 sin
,
2
2
2
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or
0
0
0
sin 30
sin
sin 30
sin
4sin
sin sin 30
,
2
or
0
0
0
0
0
30
30
30
30
2sin
cos
2sin
cos
4sin
sin sin 30
,
2
2
2
2
0
0
0
0
0
30
30
30
30
2sin
cos
2sin
cos
4sin
sin sin 30
.
(1)
2
2
2
2
From this
0
sin 30
0
and
0
sin 30
4sin sin
is taken .
1)
0
sin 30
0
from
0
30
(2)
come comes out
2)
0
sin 30
4sin sin
(3)
from being
0
30
if , is given relationship
and
of each how in value ( as
well as
,
a b
of each how in the value of ) is appropriate will be
(3) according to the formula
of value
and
of to the value of depends
has been without countless will
sin
0,
be
sin
0
from being
0
sin 30
0,
ie
0
150
will be (3) formula from exchange after :
0
sin 30
4sin sin
;
0
0
sin 30 cos
cos30 sin
4sin
sin cos
cos sin
;
2
1
3
cos
sin
4sin
sin cos
4sin
cos ;
2
2
2
cos
3 sin
8sin
sin
cos
8sin
cos ;
2
1 8sin
cos
4sin 2
3 sin
or
2
1 4 1 cos 2
1 8sin
4cos 2
3
.
(4)
4sin 2
3
4sin 2
3
4sin 2
3
tg
Fraction photo the most big or the denominator the most small was without (4)
fraction the most big to value achieves Accordingly ,
cos2
of the most big value
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cos 2
0
that it was for if we put it in (4) ,
1
2
3
tg
and
0
150
is equal to
will be
Issue
3.
In
the
triangle
0
43
when
2
2
2
sin
sin
sin sin
S
ab
relationship appropriate if , of the triangle the
rest find the angles .
This issue is the first become belongs to This problem in 2 ways we solve :
Method 1
. Triangle face :
sin
.
2
ab
S
That's a given to the relationship put and then him
ab
we reduce to :
2
2
2
sin
sin
sin
sin sin
(5)
or
2
2
sin
sin
sin
sin
sin
.
one on the left squares the difference to the lifters separate we write :
2sin
cos
2cos
sin
sin
sin
sin
2
2
2
2
or
sin
sin
sin
sin
sin
.
(6)
sin
sin
and put it in ( 6 ) :
sin
sin
sin .
the left side of this
sin
ni , right by sin
the subtract :
sin
sin
sin
sin
sin
or
2sin cos
sin
or
sin
1 2sin
0.
In this
sin
0
from being
1 2cos
0,
1
cos
.
2
But
0
180
from being
0
120
is , from which :
0
0
0
0
180
120
43
17 .
So,
0
17 .
Method 2
. (5) is
2
4
R
if we increase to ( in this case
R
external circle radius ),
2
2
2
2
2
2
2
4
sin
4
sin
4
sin
4
sin sin
R
R
R
R
become
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2
sin
sin
sin
a
b
c
R
from
2
2
2
2
2
2
2
2
2
4
,
4
,
4
sin
sin
sin
a
b
c
R
R
R
or
2
2
2
2
2
2
2
2
2
sin
,sin
,sin
4
4
4
a
b
c
R
R
R
come comes out
Let's put these in (5) :
2
2
2
2
2
2
4
4
4
2
2
c
a
b
a
b
R
R
R
R
R
or
2
2
2
.
(7)
c
a
b
ab
Then look cosines from the theorem
2
2
2
2
cos
c
a
b
ab
equation (7) .
compare :
2
cos
ab
ab
or
1
cos
,
2
so
0
120
become
0
17 .
Issue 4.
Triangle sides between
2
(8)
a
c
b
relationship there is and
2
,
(9)
2
5
tg
,
2
tg
2
tg
of functions values define
This issue is second become belongs to do it in 2 ways we solve :
Method 1.
Molweide formula by :
cos
cos
cos
sin
sin
1
2
2
2
2
2
2
2 .
(10)
sin
cos
cos
sin
sin
1
2
2
2
2
2
2
2
tg
tg
a
c
b
tg
tg
According to (8).
2
a
c
b
write that can So,
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1
2
2
2.
1
2
2
tg
tg
tg
tg
in (9).
2
tg
of value if we write :
5
1
2
2
2
2
1
5
2
tg
tg
or
5
2
2
2.
5
2
2
tg
tg
From this :
5
.
(11)
2
6
tg
Now
2
tg
the we find that for as follows to replace let's see :
1
2
2 .
(12)
2
2
1
2
2
tg
tg
tg
ctg
tg
tg
Condition according to
2
2
5
tg
is equal to . So,
1
2
2
2
5
1
2
2
tg
tg
tg
tg
from this
18
.
2
35
tg
Method 2
. Half corner tangent formula
2
2
2
2
P
b P
c
tg
P P
a
and
2
2
2
2
P
a
P
b
tg
P P
c
from
2
2
2
P
b
a
c b
tg
tg
P
a
c b
will be
From this given a must according to
2
2
1
5
2
2
3
b b
tg
b b
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or
5
2
6
tg
comes out
2
tg
and that 's it road we find
Issue 5.
If
,
and
are triangular corners if cos
,
2
cos
2
and cos
2
cuts
triangle sides to be show me
This is the third issue become belongs to become him as follows we solve :
sin
1,
2
sin
1
2
from being
0
cos
cos 90
sin
sin
cos
sin
cos
.
2
2
2
2
2
2
2
come comes out , that is one side the other 2 sides from the assembly line small
from being , this from cuts triangle to make possible come comes out
CONCLUSION
Current in the day elementary mathematics of science separately
indispensable part of trigonometry another sciences with mutually dependence ,
wide branching , science and technology development in development each in step
to trigonometry face we will come That is from mathematics except another sciences
in learning this department of science deep demands to know .
Department of trigonometry each bilaterally and deep study in this direction
addition scientific and methodical materials through mathematician knowledge and
skills expand to the goal according to will be
Geometry school in the chair another fields such as theory in trigonometry in
practice application to achieve , that is issues solve qualification it is required to
acquire . But in practice this in the field many to difficulties face will come .
Like this practical also trigometric in problems issues in solving row equivalent
of proportions application reach important role plays
REFERENCES
Books
[1 ] Novosyolov , S. I. (1954).
Special course trigonometry
. Soviet science.
[2] Rybkin N. (1933).
Collection of trigonometry
. You flew .
[3] Pogorelov A.I. (1949).
Collection of trigonometry
. Gosudarstvennoe
uchebno-pedagogicheskoe izdatelstvo Ministerstva prosveshcheniya RSFSR .
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[4] Eisenstadt Ya.I. Belotserkovskaya B. G. (1960)
Reshenie zadach po
trigonometrii.
You flew.
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