RoboguruRoboguru
SD

Diketahui △ABC, siku-siku di B, dan sin A=257​. Tentukanlah nilai tan (A−C).

Pertanyaan

Diketahui increment ABC, siku-siku di B, dan sin space straight A equals 7 over 25. Tentukanlah nilai tan space open parentheses straight A minus straight C close parentheses.

M. Ulfa

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Pembahasan

Diketahui:

increment ABC, siku-siku di B, dan sin space straight A equals 7 over 25, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight A end cell equals cell 7 over 25 end cell row cell depan over miring end cell equals cell 7 over 25 end cell end table

Digambarkan sebagai berikut:

Kemudian dengan menggunakan Teorema Pythagoras, diperoleh sisi AB yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell samping squared end cell equals cell miring squared minus depan squared end cell row blank equals cell 25 squared minus 7 squared end cell row blank equals cell 625 minus 49 end cell row blank equals 576 row samping equals cell plus-or-minus square root of 576 end cell row blank equals cell plus-or-minus 24 end cell row blank equals 24 end table

Sehingga diperoleh nilai tangen pada segitiga yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space straight A end cell equals cell fraction numerator depan space angle straight A over denominator samping space angle straight A end fraction end cell row blank equals cell 7 over 24 end cell row blank blank blank row cell tan space straight C end cell equals cell fraction numerator depan space angle straight C over denominator samping space angle straight C end fraction end cell row blank equals cell 24 over 7 end cell end table

 

Rumus tangen selisih dua sudut yaitu:

tan space open parentheses alpha minus beta close parentheses equals fraction numerator tan space alpha minus tan space beta over denominator 1 plus tan space alpha times tan space beta end fraction

Sehingga diperoleh penyelesaiannya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses straight A minus straight C close parentheses end cell equals cell fraction numerator tan space straight A minus tan space straight C over denominator 1 plus tan space straight A times tan space straight C end fraction end cell row blank equals cell fraction numerator begin display style 7 over 24 end style minus begin display style 24 over 7 end style over denominator 1 plus begin display style fraction numerator up diagonal strike 7 over denominator down diagonal strike 24 end fraction end style times begin display style fraction numerator down diagonal strike 24 over denominator up diagonal strike 7 end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 7 open parentheses 7 close parentheses over denominator 24 open parentheses 7 close parentheses end fraction end style minus begin display style fraction numerator 24 open parentheses 24 close parentheses over denominator 7 open parentheses 24 close parentheses end fraction end style over denominator 1 plus 1 end fraction end cell row blank equals cell fraction numerator begin display style 49 over 168 end style minus begin display style 576 over 168 end style over denominator 2 end fraction end cell row blank equals cell fraction numerator negative begin display style 527 over 168 end style over denominator 2 end fraction end cell row blank equals cell negative 527 over 168 times 1 half end cell row blank equals cell negative 527 over 336 end cell end table

Dengan demikiam, nilai tan space open parentheses straight A minus straight C close parentheses adalah negative 527 over 336.

38

0.0 (0 rating)

Pertanyaan serupa

Jika sin α=53​ dan cos β=178​, dengan α tumpul dan β lancip, maka tan (α−β)= ...

54

5.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia