Roboguru

Segitiga KLM memiliki koordinat . Nllai  dan   ber...

Segitiga KLM memiliki koordinat K left parenthesis negative 5 comma negative 2 right parenthesis comma space L left parenthesis 3 comma negative 2 right parenthesis comma space dan space M left parenthesis negative 5 comma 4 right parenthesis space. Nllai cos open parentheses L close parentheses space dan tan open parentheses M close parentheses  berturut-turut ... 

  1. 4 over 5 space dan space 4 over 3 space 

  2. 3 over 5 space dan space 3 over 4 space 

  3. 3 over 4 space dan space 3 over 5 space 

  4. 3 over 4 space dan space 4 over 3 space space 

  5. 3 over 5 space dan space 3 over 4 space space 

Jawaban:

Diketahui : 

Ditanya : cos open parentheses straight L close parentheses space dan space tan open parentheses straight M close parentheses ?

Jawab: 

  • Menentukan panjang K L comma space L M comma space M K space.

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar K L close vertical bar end cell equals cell open vertical bar L minus K close vertical bar end cell row blank equals cell square root of open parentheses 3 plus 5 close parentheses squared plus left parenthesis negative 2 plus 2 right parenthesis squared end root end cell row blank equals cell square root of 8 squared end root end cell row blank equals 8 row cell open vertical bar L M close vertical bar end cell equals cell open vertical bar M minus L close vertical bar end cell row blank equals cell square root of open parentheses negative 5 minus 3 close parentheses squared plus left parenthesis 4 plus 2 right parenthesis squared end root end cell row blank equals cell square root of left parenthesis negative 8 right parenthesis squared plus left parenthesis 6 right parenthesis squared end root end cell row blank equals cell square root of 64 plus 36 end root end cell row blank equals cell square root of 100 end cell row blank equals 10 row cell open vertical bar M K close vertical bar end cell equals cell open vertical bar K minus M close vertical bar end cell row blank equals cell square root of open parentheses negative 5 plus 5 close parentheses squared plus left parenthesis negative 2 minus 4 right parenthesis squared end root end cell row blank equals cell square root of left parenthesis negative 6 right parenthesis squared end root end cell row blank equals 6 end table 

  • Menentukan c o s open parentheses L close parentheses space dan space t a n open parentheses M close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row cell cos open parentheses L close parentheses end cell equals cell samping over miring end cell row blank equals cell fraction numerator K L over denominator L M end fraction 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 open parentheses M close parentheses end cell equals cell depan over samping end cell row blank equals cell fraction numerator K L over denominator M K end fraction end cell row blank equals cell 8 over 6 end cell row blank equals cell 4 over 3 end cell end table

Oleh karena itu, jawaban yang benar adalah A. 

0

Ruangguru

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