Roboguru

1.    Tentukan nilai !

Pertanyaan

1.

  

Tentukan nilai sin space straight R comma space cos space straight R comma space dan space tan space space straight Rspace space  

Pembahasan Soal:

Perhatikan penghitungan berikut!            

 sisi space miring equals square root of 1 squared plus 2 squared end root equals square root of 1 plus 4 end root equals square root of 5 sin space straight R equals fraction numerator sisi space depan over denominator sisi space miring end fraction equals fraction numerator 1 over denominator square root of 5 end fraction equals fraction numerator 1 over denominator square root of 5 end fraction cross times fraction numerator square root of 5 over denominator square root of 5 end fraction equals fraction numerator square root of 5 over denominator 5 end fraction cos space straight R equals fraction numerator sisi space samping over denominator sisi space miring end fraction equals fraction numerator 2 over denominator square root of 5 end fraction equals fraction numerator 2 over denominator square root of 5 end fraction cross times fraction numerator square root of 5 over denominator square root of 5 end fraction equals fraction numerator 2 square root of 5 over denominator 5 end fraction tan space straight R equals fraction numerator sisi space depan over denominator sisi space samping end fraction equals 1 half      

Jadi, nilai sin space straight R comma space cos space straight R comma space dan space tan space space straight R secara berturut-turut adalah 1 fifth square root of 5 comma space 2 over 5 square root of 5 comma space dan space 1 half. space 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Fatimatuzzahroh

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Tentukanlah nilai  dan  dari segitiga di bawah ini.

Pembahasan Soal:

Segitiga ABC merupakan segitiga siku-siku, maka gunakan teorema Pythagoras untuk menentukan panjang CB.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell AB squared end cell equals cell AC squared plus CB squared end cell row cell CB squared end cell equals cell AB squared minus AC squared end cell row CB equals cell plus-or-minus square root of AB squared minus AC squared end root end cell row CB equals cell plus-or-minus square root of 2 squared minus 1 squared end root end cell row CB equals cell plus-or-minus square root of 4 minus 1 end root end cell row CB equals cell plus-or-minus square root of 3 space cm end cell end table end style 

Karena panjang sisi segitiga tidak mungkin negatif, maka pilih begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row blank blank CB end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell square root of 3 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank space end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cm end table end style.

Sehingga,

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight B end cell equals cell depan over miring end cell row blank equals cell AC over AB end cell row blank equals cell 1 half end cell row cell cos space straight B end cell equals cell samping over miring end cell row blank equals cell CB over AB end cell row blank equals cell fraction numerator square root of 3 over denominator 2 end fraction end cell row blank equals cell 1 half square root of 3 end cell row cell tan space straight B end cell equals cell depan over samping end cell row blank equals cell AC over CB end cell row blank equals cell fraction numerator 1 over denominator square root of 3 end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 3 end fraction times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell 1 third square root of 3 end cell end table end style  

Dengan demikian, nilai begin mathsize 14px style sin space straight B comma space cos space straight B end style dan Error converting from MathML to accessible text. berturut-turut adalah begin mathsize 14px style 1 half comma space 1 half square root of 3 comma space 1 third square root of 3 end style.

0

Roboguru

Pada  siku-siku di B, serta panjang , tentukan: a. Panjang  b.

Pembahasan Soal:

a. Panjang AC

table attributes columnalign right center left columnspacing 0px end attributes row AC equals cell square root of AB squared plus BC squared end root end cell row blank equals cell square root of 5 squared plus 12 squared end root end cell row blank equals cell square root of 25 plus 144 end root end cell row blank equals cell square root of 169 end cell row blank equals cell 13 space cm end cell end table

b. sin space straight A comma space cos space straight A comma space dan space tan space straight A

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space A end cell equals cell fraction numerator depan space sudut space straight A over denominator miring space sudut space straight A end fraction equals 12 over 13 end cell row cell cos space A end cell equals cell fraction numerator samping space sudut space straight A over denominator miring space sudut space straight A end fraction equals 5 over 13 end cell row cell tan space A end cell equals cell fraction numerator depan space sudut space straight A over denominator samping space sudut space straight A end fraction equals 12 over 5 end cell row blank blank blank end table

0

Roboguru

Pada sebuah segitiga KLM siku-siku di  L berlaku  dan panjang sisi . Tentukan panjang sisi yang lain dan nilai perbandingan trigonometri lainnya!

Pembahasan Soal:

Diketahui

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space K end cell equals cell 1 third end cell row KL equals cell square root of 63 end cell end table  

Segitiga KLM siku-siku di L maka dapat diilustrasikan sebagai berikut.

Berdasarkan definisi tangen, maka dapat ditentukan panjang sisi LM dan KMsebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space K end cell equals cell LM over KL end cell row cell 1 third end cell equals cell fraction numerator LM over denominator square root of 63 end fraction end cell row LM equals cell 1 third square root of 63 end cell row blank equals cell 1 third square root of 9 middle dot 7 end root end cell row blank equals cell 1 third middle dot 3 square root of 7 end cell row blank equals cell square root of 7 end cell row KM equals cell square root of KL squared plus LM squared end root end cell row blank equals cell square root of open parentheses square root of 63 close parentheses squared plus open parentheses square root of 7 close parentheses squared end root end cell row blank equals cell square root of 63 plus 7 end root end cell row blank equals cell square root of 70 end cell end table

Ingat definisi trigonometri

sinxcosxcscxsecxcotx=====miringdepanmiringsampingsinx1cosx1tanx1

LM sisi depan sudut K dan KL sisi samping sudut K maka

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight K end cell equals cell LM over KM end cell row blank equals cell fraction numerator square root of 7 over denominator square root of 70 end fraction end cell row blank equals cell 1 over 10 square root of 10 end cell row cell cos space straight K end cell equals cell KL over KM end cell row blank equals cell fraction numerator square root of 63 over denominator square root of 70 end fraction end cell row blank equals cell fraction numerator 3 square root of 7 over denominator square root of 70 end fraction end cell row blank equals cell 3 over 10 square root of 10 end cell row cell cot space straight K end cell equals cell fraction numerator 1 over denominator tan space straight K end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style bevelled 1 third end style end fraction end cell row blank equals 3 row cell sec space straight K end cell equals cell fraction numerator 1 over denominator cos space straight K end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style bevelled fraction numerator 3 square root of 10 over denominator 10 end fraction end style end fraction end cell row blank equals cell fraction numerator 10 over denominator 3 square root of 10 end fraction end cell row blank equals cell 1 third square root of 10 end cell row cell cosec space straight K end cell equals cell fraction numerator 1 over denominator sin space straight K end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style bevelled fraction numerator square root of 10 over denominator 10 end fraction end style end fraction end cell row blank equals cell fraction numerator 10 over denominator square root of 10 end fraction end cell row blank equals cell square root of 10 end cell end table

Dengan demikian, panjnag sisi dan perbandingan sisi yang lain seperti pada uraian di atas. 

0

Roboguru

Nilai dari titik adalah ....

Pembahasan Soal:

Titik text P end text left parenthesis 5 comma 12 right parenthesis spacemaka x equals 5 space text dan end text space y equals 12 comma sehingga r spacedapat diperoleh sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row r equals cell plus-or-minus square root of x squared plus y squared end root end cell row r equals cell plus-or-minus square root of 5 squared plus 12 squared end root end cell row r equals cell plus-or-minus square root of 25 plus 144 end root end cell row r equals cell plus-or-minus square root of 169 end cell row r equals cell plus-or-minus 13 end cell end table


Karena r spaceselalu bernilai positif maka nilai yang tepat adalah r equals 13 comma sehingga diperoleh

text Sin end text space text P end text equals y over r equals 12 over 13 text Cos end text space text P end text equals x over r equals 5 over 13 text Tan end text space text P end text equals y over x equals 12 over 5

 

Jadi nilai dari text sin, cos, dan tan  end textdari titik text P end text left parenthesis 5 comma 12 right parenthesis spaceadalah 12 over 13 comma 5 over 13 comma 12 over 5.

0

Roboguru

Diketahui segitiga siku-siku berikut Tentukan perbandingan trigonometri segitiga di atas

Pembahasan Soal:

Diketahui segitiga ABC siku-siku di B, sehingga berlaku teorema Pythagoras sebagai berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell AB squared end cell equals cell AC squared minus BC squared end cell row blank equals cell 10 squared minus 8 squared end cell row blank equals cell 100 minus 64 end cell row blank equals 36 row blank blank blank row AB equals cell plus-or-minus 6 space cm end cell end table

Karena panjang seagitiga tidak mungkin negatif, maka yang memenuhi adalah AB equals 6 space cm. Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell AB over AC end cell row blank equals cell 6 over 10 end cell row blank equals cell 3 over 5 end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell BC over AC end cell row blank equals cell 8 over 10 end cell row blank equals cell 4 over 5 end cell end table 

 table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell AB over BC end cell row blank equals cell 6 over 8 end cell row blank equals cell 3 over 4 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell sec space alpha end cell equals cell AC over BC end cell row blank equals cell 10 over 8 end cell row blank equals cell 5 over 4 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell csc space alpha end cell equals cell AC over AB end cell row blank equals cell 10 over 6 end cell row blank equals cell 5 over 3 end cell end table

table attributes columnalign right center left columnspacing 0px end attributes row cell cot space alpha end cell equals cell BC over AB end cell row blank equals cell 8 over 6 end cell row blank equals cell 4 over 3 end cell end table

Jadi, nilai dari sin space alpha equals 3 over 5table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell 4 over 5 end cell end tabletable attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell 3 over 4 end cell end tabletable attributes columnalign right center left columnspacing 0px end attributes row cell sec space alpha end cell equals cell 5 over 4 end cell end tabletable attributes columnalign right center left columnspacing 0px end attributes row cell csc space alpha end cell equals cell 5 over 3 end cell end tabletable attributes columnalign right center left columnspacing 0px end attributes row cell cot space alpha end cell equals cell 4 over 3 end cell end table 

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved