Roboguru

Nilai dari 16sin230∘+5tan260∘−4sin245∘ adalah ....

Pertanyaan

Nilai dari begin mathsize 14px style 16 space sin squared space 30 degree plus 5 space tan squared space 60 degree minus 4 space sin squared space 45 degree end style adalah ....

  1. begin mathsize 14px style 21 end style 

  2. begin mathsize 14px style 17 end style 

  3. begin mathsize 14px style 15 end style 

  4. begin mathsize 14px style 7 end style 

  5. begin mathsize 14px style 5 end style 

Pembahasan Soal:

begin mathsize 14px style 16 space sin squared space 30 degree plus 5 space tan squared space 60 degree minus 4 sin squared space 45 degree equals 16 open parentheses 1 half close parentheses squared plus 5 open parentheses square root of 3 close parentheses squared minus 4 open parentheses fraction numerator square root of 2 over denominator 2 end fraction close parentheses squared equals 16 open parentheses 1 fourth close parentheses plus 5 open parentheses 3 close parentheses minus 4 open parentheses 2 over 4 close parentheses equals 4 plus 15 minus 2 equals 17 end style 

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

K. Putri

Mahasiswa/Alumni Universitas Pendidikan Ganesha

Terakhir diupdate 05 Oktober 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

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 sin75∘!

Pembahasan Soal:

begin mathsize 14px style sin 75 degree equals DE over AD DE equals DF plus EF end style


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin 45 degree end cell equals cell DF over BD end cell row cell fraction numerator square root of 2 over denominator 2 end fraction end cell equals cell DF over BD end cell row DF equals cell fraction numerator square root of 2 over denominator 2 end fraction BD end cell end table end style 


  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell tan 60 degree end cell equals cell AB over BD end cell row cell square root of 3 end cell equals cell AB over BD end cell row AB equals cell square root of 3 BD end cell row blank blank blank end table end style


  begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application 45 degree end cell equals cell BC over AB end cell row cell fraction numerator square root of 2 over denominator 2 end fraction end cell equals cell BC over AB end cell row BC equals cell fraction numerator square root of 2 over denominator 2 end fraction AB end cell row EF equals cell BC equals fraction numerator square root of 2 over denominator 2 end fraction AB end cell end table end style


 begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin 30 degree end cell equals cell BD over AD end cell row cell 1 half end cell equals cell BD over AD end cell row AD equals cell 2 BD end cell end table end style


begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin invisible function application 75 degree end cell equals cell DE over AD end cell row blank equals cell fraction numerator DF plus EF over denominator AD end fraction end cell row blank equals cell fraction numerator fraction numerator square root of 2 over denominator 2 end fraction BD plus fraction numerator square root of 2 over denominator 2 end fraction AB over denominator 2 BD end fraction end cell row blank equals cell fraction numerator fraction numerator square root of 2 over denominator 2 end fraction BD plus fraction numerator square root of 2 over denominator 2 end fraction bullet square root of 3 BD over denominator 2 BD end fraction end cell row blank equals cell fraction numerator fraction numerator square root of 2 over denominator 2 end fraction plus fraction numerator square root of 6 over denominator 2 end fraction over denominator 2 end fraction end cell row blank equals cell fraction numerator 1 half open parentheses square root of 2 plus square root of 6 close parentheses over denominator 2 end fraction end cell row blank equals cell 1 fourth open parentheses square root of 2 plus square root of 6 close parentheses end cell end table end style

0

Roboguru

Nilai cos210∘+tan300∘−sin120∘ adalah ....

Pembahasan Soal:

begin mathsize 14px style cos space 210 degree plus tan space 300 degree minus sin space 120 degree equals cos space open parentheses 180 degree plus 30 degree close parentheses plus tan open parentheses 360 degree minus 60 degree close parentheses minus sin open parentheses 180 degree minus 60 degree close parentheses equals negative cos space 30 degree plus open parentheses negative tan space 60 degree close parentheses minus sin space 60 degree equals negative fraction numerator square root of 3 over denominator 2 end fraction minus square root of 3 minus fraction numerator square root of 3 over denominator 2 end fraction equals negative 2 square root of 3 end style 

Jadi, jawaban yang tepat adalah A.

0

Roboguru

Diketahui tanα=43​ dan cosβ=1312​. Jika α dan β sudut lancip, nilai dari sinα+cotanβ adalah ....

Pembahasan Soal:

Dengan menggunakan triple Pythagoras maka diperoleh sisi-sisi lainnya seperti gambar di bawah ini.

Sehingga,

begin mathsize 14px style sin space straight alpha plus cotan space straight beta equals 3 over 5 plus 12 over 5 equals 15 over 5 equals 3 end style

2

Roboguru

Diketahui segitiga PQR siku-siku di Q dimana PQ=12cm dan sinR=21​3​. Panjang sisi PR+QR=...cm

Pembahasan Soal:

table attributes columnalign right center left columnspacing 0px end attributes row cell Sin space straight R end cell equals cell depan over miring end cell row cell 1 half square root of 3 end cell equals cell PQ over PR end cell row cell 1 half square root of 3 end cell equals cell 12 over PR end cell row PR equals cell fraction numerator 12 over denominator begin display style 1 half end style 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 blank equals cell fraction numerator 24 square root of 3 over denominator square root of 3 end fraction end cell row blank equals cell 8 square root of 3 space cm end cell end table

karena  sin space straight R equals 1 half square root of 3 maka besar sudur R sebagai berikut

table attributes columnalign right center left columnspacing 0px end attributes row cell arc space sin open parentheses 1 half square root of 3 close parentheses end cell equals cell 60 degree end cell row cell angle straight R end cell equals cell 60 degree end cell row blank blank blank end table

 Panjang QR dapat ditentukan dengan definisi cos R berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space straight R end cell equals cell samping over miring end cell row cell cos space 60 degree end cell equals cell QR over PR end cell row cell 1 half end cell equals cell fraction numerator QR over denominator 8 square root of 3 end fraction end cell row QR equals cell fraction numerator 8 square root of 3 over denominator 2 end fraction end cell row QR equals cell 4 square root of 3 end cell end table 

Sehingga PR plus QR

table attributes columnalign right center left columnspacing 0px end attributes row cell PR plus QR end cell equals cell 8 square root of 3 plus 4 square root of 3 space cm end cell row blank equals cell 12 square root of 3 space cm end cell end table

Jadi, jawaban yang tepat adalah A

0

Roboguru

Perhatikan gambar persegi panjang PQRS di bawah ini. Tentukan panjang sisi RS dan PS.

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin angle SPR end cell equals cell depan over miring end cell row cell sin space 60 degree end cell equals cell RS over PR end cell row cell 1 half square root of 3 end cell equals cell RS over 20 end cell row RS equals cell 1 half square root of 3 cross times 20 end cell row RS equals cell 10 square root of 3 space cm end cell end table end style

Dengan menggunakan teorema pythagoras

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell PS squared end cell equals cell PR squared minus RS squared end cell row cell PS squared end cell equals cell 20 squared minus open parentheses 10 square root of 3 close parentheses squared end cell row cell PS squared end cell equals cell 400 minus 300 end cell row cell PS squared end cell equals 100 row PS equals cell 10 space cm end cell row blank blank blank end table end style 

Jadi, panjang RS dan PS berturut-turut adalah 10 square root of 3 space cm dan 10 space cm.

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