Roboguru

Diketahui segitiga MNO, dengan panjang MN = 24 cm, besar sudut MON = 105 dan besar sudut OMN = 30, hitung panjang sisi NO.

Pertanyaan

Diketahui segitiga MNO, dengan panjang MN = 24 cm, besar sudut MON = 105degree dan besar sudut OMN = 30degree, hitung panjang sisi NO.

Pembahasan Soal:

Dengan menggunakan aturan sinus pada perbandingan sisi-sisi segitiga maka panjang sisi NO adalah sebagai berikut:

fraction numerator M N over denominator sin space 105 degree end fraction equals fraction numerator N O over denominator sin space 30 degree end fraction 

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 105 degree end cell equals cell sin open parentheses 60 degree plus 45 degree close parentheses end cell row blank equals cell sin space 60 degree times cos space 45 degree plus cos space 60 degree times sin space 45 degree end cell row blank equals cell open parentheses 1 half square root of 3 close parentheses open parentheses 1 half square root of 2 close parentheses plus open parentheses 1 half close parentheses open parentheses 1 half square root of 2 close parentheses end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell row blank equals cell 1 fourth open parentheses square root of 6 plus square root of 2 close parentheses end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator M N over denominator sin space 105 degree end fraction end cell equals cell fraction numerator N O over denominator sin space 30 degree end fraction end cell row cell fraction numerator 24 over denominator begin display style 1 fourth end style open parentheses square root of 6 plus square root of 2 close parentheses end fraction end cell equals cell fraction numerator N O over denominator begin display style 1 half end style end fraction end cell row cell fraction numerator 24 over denominator begin display style 1 fourth end style open parentheses square root of 6 plus square root of 2 close parentheses end fraction end cell equals cell N O end cell row cell N O end cell equals cell fraction numerator 24 cross times 4 over denominator open parentheses square root of 6 plus square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator 96 over denominator open parentheses square root of 6 plus square root of 2 close parentheses end fraction cross times fraction numerator open parentheses square root of 6 minus square root of 2 close parentheses over denominator open parentheses square root of 6 minus square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator 96 open parentheses square root of 6 minus square root of 2 close parentheses over denominator 6 minus 2 end fraction end cell row blank equals cell fraction numerator 96 open parentheses square root of 6 minus square root of 2 close parentheses over denominator 4 end fraction end cell row blank equals cell 24 open parentheses square root of 6 minus square root of 2 close parentheses space cm end cell end table 

Dengan demikian, panjang sisi NO adalah 24 open parentheses square root of 6 minus square root of 2 close parentheses cm.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Tessalonika

Mahasiswa/Alumni Universitas Negeri Medan

Terakhir diupdate 07 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Tentukan panjang QR!

Pembahasan Soal:

Besar sudut R dapat ditentukan dengan cara berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell angle R end cell equals cell 180 degree minus 30 degree minus 45 degree end cell row blank equals cell 105 degree end cell end table 

Sin 105 degree dapat ditentukan dengan cara berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 105 degree end cell equals cell sin open parentheses 60 degree plus 45 degree close parentheses end cell row blank equals cell sin space 60 degree cos space 45 degree plus cos space 60 degree space sin space 45 degree end cell row blank equals cell 1 half square root of 3 times 1 half square root of 2 plus 1 half times 1 half square root of 2 end cell row blank equals cell 1 fourth square root of 6 plus 1 fourth square root of 2 end cell row blank equals cell 1 fourth open parentheses square root of 6 plus square root of 2 close parentheses end cell end table 

Dengan aturan sinus akan ditentukan panjang QR sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator QR over denominator sin space 30 degree end fraction end cell equals cell fraction numerator PQ over denominator sin space 105 degree end fraction end cell row cell fraction numerator QR over denominator begin display style 1 half end style end fraction end cell equals cell fraction numerator 50 over denominator begin display style 1 fourth open parentheses square root of 6 plus square root of 2 close parentheses end style end fraction end cell row QR equals cell fraction numerator begin display style 1 half end style times 50 over denominator begin display style 1 fourth end style open parentheses square root of 6 plus square root of 2 close parentheses end fraction end cell row QR equals cell fraction numerator 25 over denominator square root of 6 plus square root of 2 end fraction cross times fraction numerator 4 open parentheses square root of 6 minus square root of 2 close parentheses over denominator open parentheses square root of 6 minus square root of 2 close parentheses end fraction end cell row blank equals cell fraction numerator 25 times 4 times open parentheses square root of 6 minus square root of 2 close parentheses over denominator 6 minus 2 end fraction end cell row blank equals cell fraction numerator 25 times up diagonal strike 4 times open parentheses square root of 6 minus square root of 2 close parentheses over denominator up diagonal strike 4 end fraction end cell row blank equals cell 25 open parentheses square root of 6 minus square root of 2 close parentheses space cm end cell end table  

Jadi, panjang QR adalah 25 open parentheses square root of 6 minus square root of 2 close parentheses cm

 

Roboguru

Perhatikan gambar berikut! Dengan menggunakan aturan Sinus dan Cosinus, tentukan panjang !

Pembahasan Soal:

Dengan menggunakan aturan Sinus:

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator AC over denominator sin space 30 degree end fraction end cell equals cell fraction numerator AB over denominator sin space angle straight C end fraction end cell row cell fraction numerator 10 cm over denominator begin display style bevelled 1 half end style end fraction end cell equals cell fraction numerator 15 cm over denominator sin space angle straight C end fraction end cell row cell sin space angle straight C end cell equals cell 15 over 10 cross times 1 half end cell row blank equals cell 3 over 4 end cell row cell angle straight C end cell equals cell 48 comma 59 degree end cell end table

Jadi, panjang BC adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell BC squared end cell equals cell AC squared plus AB squared minus 2 times AC times AB space cos space left parenthesis 48 comma 59 degree right parenthesis end cell row blank equals cell 10 squared plus 15 squared minus 2 times 10 times 15 times 3 over 4 end cell row blank equals cell 100 plus 225 minus 225 end cell row blank equals 100 row BC equals cell 10 cm end cell end table

Roboguru

Pada segitiga , jika , , . Maka carilah nilai .

Pembahasan Soal:

Dengan menggunakan aturan sinus, maka didapatkan perbandingan :

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator AC over denominator sin space B end fraction end cell equals cell fraction numerator AB over denominator sin space C end fraction end cell row cell fraction numerator 12 over denominator 4 over 5 end fraction end cell equals cell fraction numerator 1 over denominator sin space C end fraction end cell row cell sin space C end cell equals cell 4 over 5 cross times 1 over 12 end cell row cell sin space C end cell equals cell 1 over 15 end cell end table

Dari nilai sinus diketahui :

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space C end cell equals cell 1 over 15 end cell row cell depan over miring end cell equals cell 1 over 15 end cell end table

Sehingga sisi sampingnya :

table attributes columnalign right center left columnspacing 0px end attributes row samping equals cell square root of 15 squared minus 1 squared end root end cell row blank equals cell square root of 225 minus 1 end root end cell row blank equals cell square root of 224 end cell row blank equals cell 4 square root of 14 end cell end table

Maka, cosinusnya :

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space C end cell equals cell samping over miring end cell row blank equals cell fraction numerator 4 square root of 14 over denominator 15 end fraction end cell end table

Maka, table attributes columnalign right center left columnspacing 0px end attributes row cell cos space C end cell equals cell fraction numerator 4 square root of 14 over denominator 15 end fraction end cell end table

Roboguru

Maka  adalah

Pembahasan Soal:

Dengan menggunakan aturan sinus (perbandingan sisi segitiga), diperoleh:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator sin space B over denominator AC end fraction end cell equals cell fraction numerator sin space C over denominator AB end fraction end cell row cell fraction numerator sin space 30 degree over denominator 30 end fraction end cell equals cell fraction numerator sin space C over denominator 20 end fraction end cell row cell sin space C end cell equals cell fraction numerator begin display style 1 half end style over denominator 30 end fraction cross times 20 end cell row blank equals cell 1 third end cell end table end style 

Karena begin mathsize 14px style sin space C equals 1 third end style, maka nilai begin mathsize 14px style cos space C end style dapat dicari dengan memisalkan suatu segitiga siku-siku dengan salah satu sudut lainnya sebesar sudut begin mathsize 14px style straight C end style, kemudian sisi depan sudut begin mathsize 14px style straight C end style memiliki panjang begin mathsize 14px style 1 space cm end style, dan sisi miringnya memiliki panjang begin mathsize 14px style 3 space cm end style. Panjang sisi samping sudut begin mathsize 14px style straight C end style adalah sebagai berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space samping end cell equals cell square root of 3 squared minus 1 squared end root end cell row blank equals cell square root of 9 minus 1 end root end cell row blank equals cell square root of 8 end cell row blank equals cell 2 square root of 2 space cm end cell end table end style   

Sehingga,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space C end cell equals cell fraction numerator sisi space samping over denominator sisi space miring end fraction end cell row blank equals cell fraction numerator 2 square root of 2 over denominator 3 end fraction end cell end table end style 

Jadi, nilai begin mathsize 14px style cos space C end style adalah begin mathsize 14px style fraction numerator 2 square root of 2 over denominator 3 end fraction end style

Roboguru

Diketahui sisi empat ABCD seperti gambar. Panjang sisi BC adalah...

Pembahasan Soal:

Pada setiap segitiga berlaku aturan sinus: begin mathsize 12px style fraction numerator a over denominator sin A end fraction equals fraction numerator b over denominator sin begin display style B end style end fraction equals fraction numerator c over denominator sin begin display style C end style end fraction end style

Pada segitiga tersebut berlaku aturan cosinus:

  • begin mathsize 12px style a squared equals b squared plus c squared minus 2 times b times c times cos space A end style
  • begin mathsize 12px style b squared equals a squared plus c squared minus 2 times a times c times cos space B end style
  • begin mathsize 12px style c squared equals a squared plus b squared minus 2 times a times b times cos space C end style

Perhatikan gambar!

Dari aturan sinus maka pada segitiga siku-siku ADB berlaku:

begin mathsize 12px style fraction numerator B D over denominator sin 60 end fraction equals fraction numerator A D over denominator sin begin display style 30 end style end fraction  B D equals fraction numerator sin begin display style 60 end style over denominator sin begin display style 30 end style end fraction cross times A D  B D equals fraction numerator begin display style 1 half square root of 3 end style over denominator begin display style 1 half end style end fraction cross times 2 square root of 3 space c m  B D equals 6 space c m end style

Dari aturan cosinus maka pada segitiga BCD berlaku:

begin mathsize 12px style open parentheses B C close parentheses squared equals open parentheses B D close parentheses squared plus open parentheses D C close parentheses squared minus 2 times open parentheses B D close parentheses times open parentheses D C close parentheses cos space 45 degree  open parentheses B C close parentheses squared equals open parentheses 6 space c m close parentheses squared plus open parentheses 4 square root of 2 space c m close parentheses squared minus 2 times open parentheses 6 space c m close parentheses times open parentheses 4 square root of 2 space c m close parentheses open parentheses 1 half square root of 2 close parentheses  open parentheses B C close parentheses squared equals 36 space c m squared plus 32 space c m squared minus 48 space c m squared  open parentheses B C close parentheses squared equals 20 space c m squared  B C equals square root of 20 space c m squared end root  B C equals 2 square root of 5 space c m end style

Jadi, panjang sisi BC adalah begin mathsize 12px style 2 square root of 5 end style cm

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