Roboguru

Pada segitiga ABC yang siku-siku di B berlaku . Jika panjang sisi , tentukan panjang sisi AC dan BC!

Pertanyaan

Pada segitiga ABC yang siku-siku di B berlaku begin mathsize 14px style cos space straight A equals 5 over 13 end style. Jika panjang sisi begin mathsize 14px style AB equals 10 space cm end style, tentukan panjang sisi AC dan BC!

Pembahasan Soal:

Diketahui:

begin mathsize 14px style cos space straight A equals 5 over 13 end style
begin mathsize 14px style AB equals 10 space cm end style

Mencari panjang sisi AC:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight A end cell equals cell fraction numerator square root of 13 squared minus 5 squared end root over denominator 13 end fraction space end cell row blank equals cell fraction numerator square root of 169 minus 24 end root over denominator 13 end fraction end cell row blank equals cell fraction numerator square root of 144 over denominator 13 end fraction end cell row blank equals cell 12 over 13 end cell row blank blank blank row AC equals cell fraction numerator AB over denominator cos space straight A end fraction end cell row blank equals cell fraction numerator 10 over denominator open parentheses 5 over 13 close parentheses end fraction end cell row blank equals cell space 26 space cm end cell end table end style 

Mencari panjang BC:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row BC equals cell AC cross times sin space straight A end cell row blank equals cell space fraction numerator 26 over denominator begin display style 12 over 13 end style end fraction end cell row blank equals cell 24 space cm end cell end table end style 

Jadi, panjang sisi AC dan BC berturut-turut adalah begin mathsize 14px style 26 space cm space dan space 24 space cm end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

M. Mariyam

Mahasiswa/Alumni Institut Pertanian Bogor

Terakhir diupdate 12 September 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

Perhatikan gambar berikut. Tentukan panjang CB dan AB, apabila diketahui panjang !

Pembahasan Soal:

Ingatlah perbandingan sisi (trigonometri) pada segitiga siku-siku, yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell depan over miring end cell row cell cos space alpha end cell equals cell samping over miring end cell end table

Sehingga diperoleh:

  • Panjang CB

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight alpha end cell equals cell CB over AC end cell row cell cos space 30 degree end cell equals cell fraction numerator CB over denominator 12 square root of 3 space cm end fraction end cell row cell 1 half square root of 3 end cell equals cell fraction numerator CB over denominator 12 square root of 3 space cm end fraction end cell row cell 2 CB end cell equals cell 12 square root of 3 space cm cross times square root of 3 end cell row cell 2 CB end cell equals cell 36 space cm end cell row CB equals cell 36 over 2 end cell row CB equals cell 18 space cm end cell end table

  • Panjang AB

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 cell sin space 30 degree end cell equals cell fraction numerator AB over denominator 12 square root of 3 space cm end fraction end cell row cell 1 half end cell equals cell fraction numerator AB over denominator 12 square root of 3 space cm end fraction end cell row cell 2 AB end cell equals cell 12 square root of 3 space cm end cell row AB equals cell fraction numerator 12 square root of 3 space cm over denominator 2 end fraction end cell row AB equals cell 6 square root of 3 space cm end cell end table

Maka, panjang CB dan AB adalah 18 space cm dan 6 square root of 3 space cm.

Roboguru

Cermati Gambar berikut! Gambar 4.35. Kombinasi segitiga siku-siku Dengan menemukan hubungan antarsudut-sudut dan panjang sisi-sisi pada segitiga siku-siku yang ada pada gambar, hitunglah !

Pembahasan Soal:

Diperoleh dari soal sebelumnya begin mathsize 14px style A E equals fraction numerator x open parentheses square root of 6 minus square root of 2 close parentheses over denominator 2 end fraction end style dan undefined.

Perhatikan!

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos 75 degree end cell equals cell cos angle D A E end cell row blank equals cell fraction numerator A E over denominator A D end fraction end cell row blank equals cell fraction numerator fraction numerator x open parentheses square root of 6 minus square root of 2 close parentheses over denominator 2 end fraction over denominator 2 x end fraction end cell row blank equals cell fraction numerator square root of 6 minus square root of 2 over denominator 4 end fraction end cell row blank blank blank row blank blank blank row blank blank blank row blank blank blank end table end style

Jadi, begin mathsize 14px style cos 75 degree equals fraction numerator square root of 6 minus square root of 2 over denominator 4 end fraction end style.

Roboguru

Tentukan panjang AB berikut.

Pembahasan Soal:

Berdasarkan aturan trigonometri, diperoleh

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell cos open parentheses 30 degree close parentheses end cell equals cell fraction numerator sisi space samping over denominator sisi space miring end fraction end cell row cell 1 half square root of 3 end cell equals cell AB over BC end cell row cell 1 half square root of 3 end cell equals cell fraction numerator AB over denominator 10 square root of 3 end fraction end cell row cell 1 half square root of 3 cross times 10 square root of 3 end cell equals cell fraction numerator AB over denominator 10 square root of 3 end fraction cross times 10 square root of 3 end cell row 15 equals AB end table end style 

Jadi, panjang begin mathsize 14px style AB equals 15 space cm end style.

Roboguru

Perhatikan gambar berikut!     Hitunglah panjang , , , , dan !

Pembahasan Soal:

Gunakan konsep sudut istimewa trigonometri dan perbandingan sisi trigonometri.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space alpha end cell equals cell fraction numerator sisi space depan space alpha over denominator sisi space miring end fraction end cell row cell tan space alpha end cell equals cell fraction numerator sisi space depan space alpha over denominator sisi space samping space alpha end fraction end cell row cell tan space 30 degree end cell equals cell 1 third square root of 3 end cell row cell tan space 45 degree end cell equals 1 row cell sin space 30 degree end cell equals cell 1 half end cell row cell sin space 45 degree end cell equals cell 1 half square root of 2 end cell end table

*Menentukan panjang CD.

table attributes columnalign right center left columnspacing 2px end attributes row cell tan space 45 degree end cell equals cell fraction numerator sisi space depan space 45 degree over denominator sisi space damping space 45 degree end fraction end cell row 1 equals cell CD over BC end cell row 1 equals cell CD over 18 end cell row CD equals cell 18 space cm end cell end table

*Menentukan panjang AC.

table attributes columnalign right center left columnspacing 2px end attributes row cell tan space 30 degree end cell equals cell fraction numerator sisi space depan space 30 degree over denominator sisi space damping space 30 degree end fraction end cell row cell 1 third square root of 3 end cell equals cell CD over AC end cell row cell 1 third square root of 3 end cell equals cell 18 over AC end cell row AC equals cell fraction numerator 18 over denominator begin display style 1 third end style square root of 3 end fraction end cell row AC equals cell fraction numerator 18 over denominator begin display style fraction numerator square root of 3 over denominator 3 end fraction end style end fraction end cell row AC equals cell 18 cross times fraction numerator 3 over denominator square root of 3 end fraction end cell row AC equals cell fraction numerator 54 over denominator square root of 3 end fraction cross times fraction numerator square root of 3 over denominator square root of 3 end fraction end cell row AC equals cell 54 over 3 square root of 3 end cell row AC equals cell 18 square root of 3 space cm end cell end table

*Menentukan panjang AD.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space 30 degree end cell equals cell fraction numerator sisi space depan space 30 degree over denominator sisi space miring end fraction end cell row cell 1 half end cell equals cell CD over AD end cell row cell 1 half end cell equals cell 18 over AD end cell row cell AD cross times 1 end cell equals cell 18 cross times 2 end cell row AD equals cell 36 space cm end cell end table

*Menentukan panjang BD.

table attributes columnalign right center left columnspacing 2px end attributes row cell sin space 45 degree end cell equals cell fraction numerator sisi space depan space 45 degree over denominator sisi space miring end fraction end cell row cell 1 half square root of 2 end cell equals cell CD over BD end cell row cell 1 half square root of 2 end cell equals cell 18 over BD end cell row BD equals cell fraction numerator 18 over denominator begin display style 1 half end style square root of 2 end fraction end cell row BD equals cell fraction numerator 18 over denominator begin display style fraction numerator square root of 2 over denominator 2 end fraction end style end fraction end cell row BD equals cell 18 cross times fraction numerator 2 over denominator square root of 2 end fraction end cell row BD equals cell fraction numerator 36 over denominator square root of 2 end fraction cross times fraction numerator square root of 2 over denominator square root of 2 end fraction end cell row BD equals cell 36 over 2 square root of 2 end cell row BD equals cell 18 square root of 2 space cm end cell end table

*Menemukan panjang AB.

table attributes columnalign right center left columnspacing 2px end attributes row AC equals cell AB plus BC end cell row cell 18 square root of 3 end cell equals cell AB plus 18 end cell row AB equals cell open parentheses 18 square root of 3 minus 18 close parentheses space cm end cell end table

Jadi, diperoleh panjang CDACADBD, dan AB berturut-turut adalah 18 space cm18 square root of 3 space cm36 space cm18 square root of 2 space cm, dan open parentheses 18 square root of 3 minus 18 close parentheses space cm.

Roboguru

Sebuah segitiga PQR siku-siku di P dengan panjang  cm dan sudut . Panjang QR adalah ... cm

Pembahasan Soal:

Diketahui:
segitiga PQR siku-siku di P
PQ space left parenthesis Sisi space samping space sudut right parenthesis equals 4
angle straight Q equals 30 degree

Ditanyakan:
Q R space open parentheses Sisi space miring close parentheses equals... ? 

Penyelesaian:
Perbandingan sisi pada trigonometri untuk cosinus yaitu:

cos space straight alpha equals samping over miring

Dengan menggunakan perbandingan sisi pada trigonometri, maka berlaku perbandingan:

table attributes columnalign right center left columnspacing 0px end attributes row cell Tan space 30 degree end cell equals cell samping over miring end cell row cell fraction numerator 1 over denominator square root of 3 end fraction end cell equals cell 4 over QR end cell row QR equals cell 4 square root of 3 space cm end cell end table

Jadi, panjang QR adalah table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 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.

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