Roboguru

Tentukan nilai-nilai perbandingan trigonometri pada titik berikut! a.

Pertanyaan

Tentukan nilai-nilai perbandingan trigonometri pada titik berikut!

a. straight A left parenthesis negative 3 comma 4 right parenthesis 

Pembahasan Soal:

Diketahui titik straight A left parenthesis negative 3 comma 4 right parenthesis artinya x equals negative 3 space dan space y equals 4, maka dapat kita cari panjang sisi miring left parenthesis r right parenthesis dengan menggunakan pythagoras.

table attributes columnalign right center left columnspacing 0px end attributes row r equals cell square root of x squared plus y squared end root end cell row blank equals cell square root of left parenthesis negative 3 right parenthesis squared plus 4 to the power of 2 end exponent end root end cell row blank equals cell square root of 9 plus 16 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table 

Akan ditentukan sinus dari A

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight A end cell equals cell y over r end cell row blank equals cell 4 over 5 end cell end table 

Jadi, nilai sinus dari A adalah 4 over 5.

Akan ditentukan cosinus dari A

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight A end cell equals cell x over r end cell row blank equals cell fraction numerator negative 3 over denominator 5 end fraction end cell row blank equals cell negative 3 over 5 end cell end table 

Jadi, nilai cosinus dari A adalah negative 3 over 5.

Akan ditentukan tangen dari A

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space straight A end cell equals cell y over x end cell row blank equals cell fraction numerator 4 over denominator negative 3 end fraction end cell row blank equals cell negative 4 over 3 end cell end table 

Jadi, nilai tangen dari A adalah negative 4 over 3.

Akan ditentukan cosecan dari A

table attributes columnalign right center left columnspacing 0px end attributes row cell csc space straight A end cell equals cell r over y end cell row blank equals cell 5 over 4 end cell end table 

Jadi, nilai cosecan dari A adalah 5 over 4.

Akan ditentukan secan dari A

table attributes columnalign right center left columnspacing 0px end attributes row cell sec space straight A end cell equals cell r over x end cell row blank equals cell fraction numerator 5 over denominator negative 3 end fraction end cell row blank equals cell negative 5 over 3 end cell end table 

Jadi, nilai secan dari A adalah negative 5 over 3.

Akan ditentukan cotangen dari A

table attributes columnalign right center left columnspacing 0px end attributes row cell cot space straight A end cell equals cell x over y end cell row blank equals cell fraction numerator negative 3 over denominator 4 end fraction end cell row blank equals cell negative 3 over 4 end cell end table 

Jadi, nilai cotangen dari A adalah negative 3 over 4.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

E. Lestari

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 04 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

Diketahui segitiga PQR siku siku di Q, diketahui panjang PR 5, panjang  tentukan g.

Pembahasan Soal:

Perhatikan gambar di bawah ini

Diketahui panjang PR equals 5 dan PQ equals 3. Berdasarkan teorema Pythagoras, dapat diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row QR equals cell square root of PR squared minus PQ squared end root end cell row blank equals cell square root of 5 squared minus 3 squared end root end cell row blank equals cell square root of 25 minus 9 end root end cell row blank equals cell square root of 16 end cell row blank equals 4 end table

Jadi, diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell sec space straight R end cell equals cell fraction numerator 1 over denominator cos space straight R end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style QR over PR end style end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style 4 over 5 end style end fraction end cell row blank equals cell 5 over 4 end cell end table 

0

Roboguru

Perhatikan segitiga berikut, kemudian tentukan nilai  dan !

Pembahasan Soal:

Kita cari terlebih dahulu sisi samping sudut dengan menggunakan teorema Pythagoras

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sisi space samping end cell equals cell sisi space tegak equals square root of 29 squared minus 20 squared end root end cell row blank equals cell square root of 841 minus 400 end root equals square root of 441 equals 21 end cell end table end style 

Kemudian kita tentukan  nilai begin mathsize 14px style sin space straight alpha comma space cos space straight alpha comma space tan space straight alpha comma space sec space straight alpha comma space cosec space straight alpha end style dan begin mathsize 14px style cotan space straight alpha end style.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight alpha end cell equals cell fraction numerator sisi space depan space sudut over denominator sisi space miring end fraction equals 20 over 29 end cell row cell cos space straight alpha end cell equals cell fraction numerator sisi space samping space sudut over denominator sisi space miring end fraction equals 21 over 29 end cell row cell tan space straight alpha end cell equals cell fraction numerator sisi space depan space sudut over denominator sisi space samping space sudut end fraction equals 20 over 21 end cell row cell sec space straight alpha end cell equals cell fraction numerator 1 over denominator cos space straight alpha end fraction equals 29 over 21 end cell row cell cosec space straight alpha end cell equals cell fraction numerator 1 over denominator sin space straight alpha end fraction equals 29 over 20 end cell row cell cotan space straight alpha end cell equals cell fraction numerator 1 over denominator tan space straight alpha end fraction equals 21 over 20 end cell end table end style 

 

1

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

Tentukan  perbandingan yang lain jika diketahui !

Pembahasan Soal:

Diketahui cos space beta equals q. Dengan menggunakan definisi cosinus maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space beta end cell equals q row cell cos space beta end cell equals cell q over 1 end cell row cell fraction numerator sisi space samping over denominator sisi space miring end fraction end cell equals cell q over 1 end cell row blank blank blank end table

Kemudian, dengan menggunakan teorema pythagoras diperoleh sisi depan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses sisi space depan close parentheses squared end cell equals cell open parentheses sisi space miring close parentheses squared minus open parentheses sisi space samping close parentheses squared end cell row blank equals cell 1 squared minus q squared end cell row blank equals cell 1 minus q squared end cell row cell sisi space depan end cell equals cell square root of 1 minus q squared end root end cell end table

Berdasarkan uraian di atas, maka diperoleh  5 perbandingan trigonometri yang lain sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space beta end cell equals cell fraction numerator sisi space depan over denominator sisi space miring end fraction end cell row blank equals cell fraction numerator square root of 1 minus q squared end root over denominator 1 end fraction end cell row blank equals cell square root of 1 minus q squared end root end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell tan space beta end cell equals cell fraction numerator sisi space depan over denominator sisi space samping end fraction end cell row blank equals cell fraction numerator square root of 1 minus q squared end root over denominator q end fraction end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell sec space straight beta end cell equals cell fraction numerator 1 over denominator cos space straight beta end fraction end cell row blank equals cell 1 over straight q end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell cosec space straight beta end cell equals cell fraction numerator 1 over denominator sin space straight beta end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 1 minus straight q squared end root end fraction end cell row blank equals cell fraction numerator 1 over denominator square root of 1 minus straight q squared end root end fraction cross times fraction numerator square root of 1 minus straight q squared end root over denominator square root of 1 minus straight q squared end root end fraction end cell row blank equals cell fraction numerator square root of 1 minus straight q squared end root over denominator 1 minus straight q squared end fraction end cell end table


table attributes columnalign right center left columnspacing 0px end attributes row cell cot space straight beta end cell equals cell fraction numerator 1 over denominator tan space straight beta end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator square root of 1 minus straight q squared end root over denominator straight q end fraction end style end fraction end cell row blank equals cell fraction numerator straight q over denominator square root of 1 minus straight q squared end root end fraction end cell row blank equals cell fraction numerator straight q over denominator square root of 1 minus straight q squared end root end fraction cross times fraction numerator square root of 1 minus straight q squared end root over denominator square root of 1 minus straight q squared end root end fraction end cell row blank equals cell fraction numerator straight q square root of 1 minus straight q squared end root over denominator 1 minus straight q squared end fraction end cell end table


Dengan demikian, 5 perbandingan trigonometri yang lain yaitu sin space beta equals square root of 1 minus q squared end roottan space beta equals fraction numerator square root of 1 minus q squared end root over denominator q end fractionsec space beta equals 1 over qcosec space beta equals fraction numerator square root of 1 minus q squared end root over denominator 1 minus q squared end fraction, dan c o t space beta equals fraction numerator q square root of 1 minus q squared end root over denominator square root of 1 minus q squared end root end fraction.

0

Roboguru

Diketahui , hitunglah , , , , dan !

Pembahasan Soal:

Karena diketahui begin mathsize 14px style cos space alpha end style  maka.

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight alpha end cell equals cell samping over miring end cell row blank equals cell p over 2 end cell end table end style 

Sebelum mencari yang ditanyakan, terlebih dahulu kita menentukan panjang sisi depan sudut, yaitu dengan menggunakan teorema phytagoras:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of 2 squared minus p squared end root end cell row blank equals cell square root of 4 minus p squared end root end cell end table end style

Jadi, panjang sisi depan sudut adalah begin mathsize 14px style square root of 4 minus p squared end root end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight alpha end cell equals cell depan over miring end cell row blank equals cell fraction numerator square root of 4 minus straight p squared end root over denominator 2 end fraction end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan space straight alpha end cell equals cell depan over samping end cell row blank equals cell fraction numerator square root of 4 minus straight p squared end root over denominator p end fraction end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell s e c space straight alpha end cell equals cell 1 over cos end cell row blank equals cell fraction numerator 1 over denominator begin display style p over 2 end style end fraction end cell row blank equals cell 2 over p end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cosec space straight alpha end cell equals cell 1 over sin end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator square root of 4 minus p squared end root over denominator 2 end fraction end style end fraction end cell row blank equals cell fraction numerator 2 over denominator square root of 4 minus p squared end root end fraction end cell end table end style

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cotan space straight alpha end cell equals cell 1 over tan end cell row blank equals cell fraction numerator 1 over denominator begin display style fraction numerator square root of 4 minus p squared end root over denominator p end fraction end style end fraction end cell row blank equals cell fraction numerator p over denominator square root of 4 minus p squared end root end fraction end cell end table end style

 

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