Mencari Formula Kiraan Sudut & Tinggi Segi Tiga

Page 1 of 1

You cannot start a new topic
You cannot reply to this topic

Mencari Formula Kiraan Sudut & Tinggi Segi Tiga Rate Topic:

#1 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 03 November 2008 - 11:38 AM

aku tak tau kat mana tempat nak mengadu... tp aku rasa kat sini yg paling sesuai tuk selesaikan masalah nie...

harap member2 dpt membantu dalam mencari jawapan/ formula nie.....
dulu masa blaja bleh buat tp skang aku dah lupa lansung nak buat..... klu korang bleh ajarkan step by step lagi bagus....

#2 mbek

Leftenan Muda

Group: Ahli
Posts: 783
Joined: 25-July 03
Gender:Male
Location:muwo
Interests:hehehehee
Kepakaran:Web Development (perl+php+mysql...)
Freelance:Ya

Posted 03 November 2008 - 12:14 PM

kalau ikut gambarajah kat atas tu... seingat aku...

kalau ikut Teoram Pitagoras

BC² = AB² + AC²

untuk dapatkan sudut lak leh pakai...

tan = betentangan/bersebelahan

sin = betentangan/hypotonous

kos = bersebelahan/hypotonous

hehehhee... sory le... lama dah x buat... kalau salah tlg betulkan

#3 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 03 November 2008 - 01:55 PM

QUOTE (mbek @ Nov 3 2008, 12:14 PM) <{POST_SNAPBACK}>

OK aku cuba buat BC² = AB(6.5)² + AC(13)²
BC = 42.24 + 169, BC(211.25) punca kuasa dua
BC = 14.53"
btul ke nie jwpn die....

yg bawah tu aku tak reti la pulak.... sesapa bleh xplain kan lagi....

#4 mbek

Leftenan Muda

Group: Ahli
Posts: 783
Joined: 25-July 03
Gender:Male
Location:muwo
Interests:hehehehee
Kepakaran:Web Development (perl+php+mysql...)
Freelance:Ya

Posted 03 November 2008 - 03:02 PM

BC(13)² = AB(6.5)² + AC²
169 = 42.5 + AC²
169 - 42.5 = AC²
AC² = 126.75
AC = puncakuasa(126.75)
AC = 11.258

**kalau salah tlg bgtau..

#5 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 04 November 2008 - 09:25 AM

QUOTE (mbek @ Nov 3 2008, 03:02 PM) <{POST_SNAPBACK}>

BC(13)² = AB(6.5)² + AC²
169 = 42.5 + AC²
169 - 42.5 = AC²
AC² = 126.75
AC = puncakuasa(126.75)
AC = 11.258

**kalau salah tlg bgtau..

aku salah buat la...jwpn die AC=11.25"(nie yg btul)
skang nie cam mana nak cari sudut plaks....

#6 slaughters

Leftenan Muda

Group: Ahli
Posts: 910
Joined: 28-September 06
Gender:Male
Location:Padang Batu
Kepakaran:Biase2 je
Freelance:Ya

Posted 04 November 2008 - 09:28 AM

aku minat betul soalan nie time skolah2 dulu..

#7 slaughters

Leftenan Muda

Group: Ahli
Posts: 910
Joined: 28-September 06
Gender:Male
Location:Padang Batu
Kepakaran:Biase2 je
Freelance:Ya

Posted 04 November 2008 - 09:34 AM

aku kalo jumpe soalan camnie, aku suka terbalikkan die jadik camnie..

so senang nak tengok..

ok, kita da tau sudut A = 90, B dan C tak tau, so B + C = 90.

selebihnya.. kira a ikut perkiraan mbek buat.. actually aku pun da lupe.. Hahahahahahahah..

adoi.. tapi patutnya senang je.

#8 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 04 November 2008 - 11:45 AM

QUOTE (slaughters @ Nov 4 2008, 09:34 AM) <{POST_SNAPBACK}>

aku kalo jumpe soalan camnie, aku suka terbalikkan die jadik camnie..

nak kate senang pong xgak nie... mmg btul a=90, b? & c?, b+c=90.
masalahnye baru hanya tau a jer... jadi cam mana nak dpt b or c.....
Tan, Sin, Kos. cam mana nak guna tu....???? apa formula nak cari sudut nie....

(hmmm nampakny trtangguh lagi projek aku)

#9 slaughters

Leftenan Muda

Group: Ahli
Posts: 910
Joined: 28-September 06
Gender:Male
Location:Padang Batu
Kepakaran:Biase2 je
Freelance:Ya

Posted 04 November 2008 - 12:06 PM

huhuhuhu..

dengan cara ni ko nak blajar mesti lambat gile..

kalo dalam 1 segitga, sudut dalaman mesti 180 darjah..

cube lihat bentuk segi tiga tepat. satu 90 darjah, dari dua sudut mesti 45 darjah menjadikan semua 180 darjah.

logiknya disitu, biarpun bentuk segitiga tu berubah macamana sekalipun, tetap keseluruhan sudut dalam 180 darjah.

berbeza pulak kalo segi empat tepat dan lain2.

stadi !

#10 mbek

Leftenan Muda

Group: Ahli
Posts: 783
Joined: 25-July 03
Gender:Male
Location:muwo
Interests:hehehehee
Kepakaran:Web Development (perl+php+mysql...)
Freelance:Ya

Posted 04 November 2008 - 12:13 PM

kalau ikut gambarajah tu... kita gunakan sin sebab bertentangan dgn hypotonous dah diberikan.

sin = betentangan/hypotonous
sin C = 6.5 / 13
sin C = 0.5

utk dpt berapa degree tu kene tgk buku sifir empat angka yg kita guna pakai masa sekolah2 dulu tu... aku x tau apa nama in english... cari x jumpa kat net... hahahha

tapi paling aku ingat sin 0.5 = 30^o

so sudut C = 30^o

oleh sebab itu.

sudut B = 180 - 90 - 30
sudut B = 60^o

#11 slaughters

Leftenan Muda

Group: Ahli
Posts: 910
Joined: 28-September 06
Gender:Male
Location:Padang Batu
Kepakaran:Biase2 je
Freelance:Ya

Posted 04 November 2008 - 12:26 PM

jangan sampai tersalah baca hypotonous jadi hippopotamus sudah...

#12 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 04 November 2008 - 02:36 PM

QUOTE (slaughters @ Nov 4 2008, 12:06 PM) <{POST_SNAPBACK}>

yes bro aku cube buat cara ko.... dapt jawapan lbh kurang cam mbek....
b = 60, c = 30....
tp adakah setiap sudut nie 60, 30, 90 menjadikan die 180 utk sume sudut.....
logik ke??? aku konpius

mbek: mmg btul pon masa skolah dulu pkai buku sifir 4angka.... tp masalahnya aku pon dah tak ingat nak pakai cam mana.....
tp papehal pon time kasih.... kerena membantu... aku amat menghargainya.... thanks gak kat slaughters..... harap msalah selesai.....

#13 slaughters

Leftenan Muda

Group: Ahli
Posts: 910
Joined: 28-September 06
Gender:Male
Location:Padang Batu
Kepakaran:Biase2 je
Freelance:Ya

Posted 04 November 2008 - 02:49 PM

segi tiga membawa kepada sudut dalam 180 darjah, tak kisah shape die camner pun..

kalo segi empat pulak sudut dalam 360 darjah.. tak kisah jugak shape die macamane..

p/s: ramai member2 aku dulu keliru bile segitiga and segi empat ni diletakkan dalam bulatan... padahal same je..

#14 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 04 November 2008 - 04:16 PM

QUOTE (proton4tutor @ Nov 4 2008, 02:36 PM) <{POST_SNAPBACK}>

ADA 1 lg masalah... aku cube solve kan tapi ttp xdpt jwapan la... aku nie mayb kene gune yg mbek ckap tu.... AB= 6.5, BC= 15, AC= 13.5,,,, bleh ke dapt c = 30, b=60

#15 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 06 November 2008 - 09:47 AM

QUOTE (proton4tutor @ Nov 4 2008, 04:16 PM) <{POST_SNAPBACK}>

ADA 1 lg masalah... aku cube solve kan tapi ttp xdpt jwapan la... aku nie mayb kene gune yg mbek ckap tu.... AB= 6.5, BC= 15, AC= 13.5,,,, bleh ke dapt c = 30, b=60

mintak tlong lar sangat2.... cikgu2 skalian.... please..... aku tgh buntu memikirkan masalah nie....

#16 ikanez

Kadet

Group: Ahli Biasa
Posts: 25
Joined: 08-February 08
Kepakaran:nil
Freelance:Tidak

Posted 08 November 2008 - 11:38 AM

jalan kirenye same je ngn dgn yg ini (bile aku kate same tu, bkan bermaksud jawapan die same, jalan kire jek. coz kalo ikut length baru yg ko kasi tu, mmg tak akan dpt sudut 30,60 rasenye. )

QUOTE

kalau ikut gambarajah tu... kita gunakan sin sebab bertentangan dgn hypotonous dah diberikan.

sin = betentangan/hypotonous
sin C = 6.5 / 13
sin C = 0.5

utk dpt berapa degree tu kene tgk buku sifir empat angka yg kita guna pakai masa sekolah2 dulu tu... aku x tau apa nama in english... cari x jumpa kat net... hahahha

tapi paling aku ingat sin 0.5 = 30o

so sudut C = 30o

oleh sebab itu.

sudut B = 180 - 90 - 30
sudut B = 60o

- mbek

tapi kalo malas nak bwat kire2 n nak tau jwpan direct, u can use the calculator in this website.

QUOTE

http://www.1728.com/pythgorn.htm

diharap itu dpt membantu

This post has been edited by ikanez: 08 November 2008 - 11:41 AM

#17 root

Kapten

Group: Ahli Professional
Posts: 2,092
Joined: 04-July 02
Gender:Male
Location:Perantauan Jalur Lebar Tanpa Wayar.
Interests:Komputer dan membaca
Kepakaran:./configure && make && make install
Freelance:Tidak

Posted 10 November 2008 - 05:17 PM

QUOTE (proton4tutor @ Nov 6 2008, 09:47 AM) <{POST_SNAPBACK}>

ADA 1 lg masalah... aku cube solve kan tapi ttp xdpt jwapan la... aku nie mayb kene gune yg mbek ckap tu.... AB= 6.5, BC= 15, AC= 13.5,,,, bleh ke dapt c = 30, b=60
Mintak tlong lar sangat2.... cikgu2 skalian.... please..... aku tgh buntu memikirkan masalah nie....

Anda kena guna formula hukum kosain.

a^2 = b^2 + c^2 - 2*b*c*cos(A)

Jadi untuk cari sudut A

cos(A)=(b^2+c^2-a^2)/(2*b*c)
cos(A)=(13.5^2 + 6.5^2 - 15^2 )/(2*6.5*13.5)
Maka sudut A = 90.1768391064 darjah

Maka lihatlah, sudut yang sepatutnya 90 darjah bukan lagi 90 darjah.Untuk cari sudut yang lain pula, anda laukan serupa untuk sempadan lain pula...

cos(

=(b^2-a^2-c^2)/(2*a*c)

B= 25.67917681 darjah ... bukanya 30 darjah lagi.

Akhir sekali saya singkat terus..

C= acos((6.5^2+15^2-13.5^2)/(2*6.5*15)) = 64.1575871262 darjah ... bukan lagi 60 darjah.

Rujukan:
http://en.wikipedia..../Law_of_cosines

This post has been edited by root: 10 November 2008 - 05:30 PM

#18 proton4tutor

Kadet

Group: Ahli Biasa
Posts: 11
Joined: 02-November 08
Freelance:Ya

Posted 11 November 2008 - 10:34 AM

QUOTE (root @ Nov 10 2008, 05:17 PM) <{POST_SNAPBACK}>

time kasih skang nie aku dah agak paham ngan kos sin dan sume tu... lagi 1 web wikipedia ngan yang lagi 1 pon amat berguna.... time kasih skali lagi....

#19 are_zn

Kadet

Group: Ahli Biasa
Posts: 2
Joined: 10-December 08
Freelance:Tidak

Posted 09 January 2009 - 10:29 PM

kalo gune sin ok gak...

A/sin a = B/sin b = C/sin c

#20 BudakNakal

Kadet

Group: Ahli Biasa
Posts: 23
Joined: 31-July 08
Kepakaran:Biasa2 jer
Freelance:Ya

Post icon Posted 10 January 2009 - 12:04 AM

QUOTE (proton4tutor @ Nov 3 2008, 11:38 AM) <{POST_SNAPBACK}>

Assalamualaikum.. aku torai la jawab soklan nie yer..macam menarik jer..huhuhu..

mula² kita tau AC = a, BC = 13", AB = 6.5"

Guna formula BC² = AC² + AB²

So, AC² = BC² - AB²

masukkan nilai AC² = 13² - 6.5²

AC² = 169 - 42.25
= 126.75

AC = 11.2583" tapi AC = a

So, a = 11.2583"

Soklan ke 2 - nak cari b? Semua leh guna, sin ke, kos ke, tangen pun leh..

So, better guna nilai yang dia bagi..kalo x der, guna nilai yg kita dapat pun bole..

Yang kita ader AC = 11.2583", BC = 13" dan AB = 6.5"

Kita guna Kos b = bersebelahan/hipotenus

Kos b = AB/BC

= 6.5/13

= 0.5

b = kos-1 (0.5)

= 60°

Soklan ke 3, nak carik c lak..yang ni senang kot..

Dalam sesebuah segitiga, jumlah sudut dalaman adalah bersamaan 180°

Maksudnya, b + c + 90° = 180°

Kita tau b = 60°

So, 60° + c + 90° = 180°

c + 150° = 180°

c = 180° - 150°

c = 30°

Ok.. tu sajer untuk kali nie..selamat mencoba...huhuhuhu

#21 fatah

Sarjan Mejar

Group: Ahli
Posts: 242
Joined: 26-September 06
Gender:Male
Location:Kota Damansara
Kepakaran:Information Security, Networking, Embedded, Software Engineer
Freelance:Tidak

Posted 16 February 2009 - 01:36 AM

Salah satu contoh kod yang diaplikasikan dari trigonometri

CODE

Public Function AddDynamicLineLabel(p_Extents As MapWinGIS.Extents, sLayer As String, sField As String, bNoDuplicate As Boolean) As Boolean
    Dim p_Shapefile As MapWinGIS.Shapefile
    Dim i As Long, j As Long, k As Long, handle As Long
    Dim selectedShapes As Variant
    Dim strLabel As String
    Dim Stack() As StackData
    Dim Data As StackData
    Dim yLen As Double, xLen As Double
    Dim xMin As Double, yMin As Double, zMin As Double, xMax As Double, yMax As Double, zMax As Double

    Dim MidPointIndex As Long
    Dim x1 As Double, x2 As Double, y1 As Double, y2 As Double, Angle As Double

    ReDim Stack(0)

    ' TODO: to differentiate shape type

    'Get the minimum and maximum values of x and y that define the map extents
    p_Extents.GetBounds xMin, yMin, zMin, xMax, yMax, zMax

    exitLastLoop = False

    'Get the handle of the layer
    handle = getLyrHandle(sLayer)
    'Get the shapefile in the specified layer
    Set p_Shapefile = m_Map.GetObject(handle)
    If p_Shapefile Is Nothing Then Exit Function

    If p_Shapefile.SelectShapes(p_Extents, 0, MapWinGIS.SelectMode.INTERSECTION, selectedShapes) Then
        For i = 0 To UBound(selectedShapes)
            If exitLastLoop = True Then GoTo LastLoop
            If m_Map.ShapeVisible(handle, selectedShapes(i)) = False Then GoTo NextRecordFirstLoop
            j = getFieldIndex(handle, sField)
            If j < 0 Then GoTo LastLoop
            strLabel = p_Shapefile.CellValue(j, selectedShapes(i))
            If Len(strLabel) = 0 Then GoTo NextRecordFirstLoop
            MidPointIndex = p_Shapefile.Shape(selectedShapes(i)).numPoints / 2
            x1 = p_Shapefile.QuickPoint(selectedShapes(i), MidPointIndex - 1).x
            x2 = p_Shapefile.QuickPoint(selectedShapes(i), MidPointIndex).x
            y1 = p_Shapefile.QuickPoint(selectedShapes(i), MidPointIndex - 1).y
            y2 = p_Shapefile.QuickPoint(selectedShapes(i), MidPointIndex).y
            If x2 - x1 <> 0 Then
                Angle = (Atn((y2 - y1) / (x2 - x1)) * 180) / 3.14159265358979
            Else
                Angle = 90
            End If
            If bNoDuplicate = True Then
                For k = 1 To UBound(Stack)
                    If exitLastLoop = True Then GoTo LastLoop
                    Data = Stack(k)
                    If LCase(Data.Label) = LCase(strLabel) Then
                        GoTo NextRecordFirstLoop
                    End If
                Next
            End If
            ReDim Preserve Stack(UBound(Stack) + 1)
            Data.Label = strLabel
            Stack(UBound(Stack)) = Data
            If (x1 + x2) / 2 > xMin And (x1 + x2) / 2 < xMax And (y1 + y2) / 2 > yMin And (y1 + y2) / 2 < yMax Then
                Call m_Map.AddLabelEx(handle, strLabel, RGB(0, 0, 0), (x1 + x2) / 2, (y1 + y2) / 2, MapWinGIS.tkHJustification.hjCenter, Angle)
            End If
NextRecordFirstLoop:
        Next
        AddDynamicLineLabel = True
    End If
LastLoop:
    Erase Stack
    exitLastLoop = True
End Function

Maaf, koding rojak sikit. Baru nak siapkan framework.