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

Pertanyaan

Jika sin space alpha equals 3 over 5 dan cos space beta equals 8 over 17, dengan alpha tumpul dan beta lancip, maka tan space open parentheses alpha minus beta close parentheses equals ...

  1. 6 6 over 13

  2. negative 6 6 over 13

  3. 36 over 77

  4. negative 36 over 77

  5. 77 over 36

M. Ulfa

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah A.

Pembahasan

sin space alpha equals 3 over 5 dengan alpha tumpul, maka sudut berada di kuadran II dengan sumbu-x bernilai negatif sedangkan sumbu-y bernilai positif.

table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell 3 over 5 end cell row cell y over r end cell equals cell 3 over 5 end cell end table

Dengan menggunakan Teorema Pythagoras, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared end cell equals cell r squared minus y squared end cell row blank equals cell 5 squared minus 3 cubed end cell row blank equals cell 25 minus 9 end cell row blank equals 16 row x equals cell plus-or-minus square root of 16 end cell row blank equals cell plus-or-minus 4 end cell row blank equals 4 end table

Maka,

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

 

cos space beta equals 8 over 17 dengan beta lancip, maka sudut berada di kuadran I dengan sumbu-x dan sumbu-y bernilai positif.

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space beta end cell equals cell 8 over 17 end cell row cell x over r end cell equals cell 8 over 17 end cell end table

Dengan menggunakan Teorema Pythagoras, diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell y squared end cell equals cell r squared minus x squared end cell row blank equals cell 17 squared minus 8 squared end cell row blank equals cell 289 minus 64 end cell row blank equals 225 row y equals cell plus-or-minus square root of 225 end cell row blank equals cell plus-or-minus 15 end cell row blank equals 15 end table

Maka,

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space beta end cell equals cell y over x end cell row blank equals cell 15 over 8 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

Diperoleh penyelesaiannya yaitu:

table attributes columnalign right center left columnspacing 0px end attributes row cell tan space open parentheses alpha minus beta close parentheses end cell equals cell fraction numerator tan space alpha minus tan space beta over denominator 1 plus tan space alpha times tan space beta end fraction end cell row blank equals cell fraction numerator negative begin display style 3 over 4 end style minus begin display style 15 over 8 end style over denominator 1 plus open parentheses negative begin display style 3 over 4 end style close parentheses times begin display style 15 over 8 end style end fraction end cell row blank equals cell fraction numerator negative begin display style fraction numerator 3 open parentheses 2 close parentheses over denominator 4 open parentheses 2 close parentheses end fraction end style minus begin display style 15 over 8 end style over denominator 1 minus begin display style 45 over 32 end style end fraction end cell row blank equals cell fraction numerator negative begin display style 6 over 8 end style minus begin display style 15 over 8 end style over denominator begin display style 32 over 32 end style minus begin display style 45 over 32 end style end fraction end cell row blank equals cell fraction numerator negative begin display style 21 over 8 end style over denominator negative begin display style 13 over 32 end style end fraction end cell row blank equals cell negative 21 over 8 times open parentheses negative 32 over 13 close parentheses end cell row blank equals cell 21 over 13 times 4 end cell row blank equals cell 64 over 13 end cell row blank equals cell 6 6 over 13 end cell end table

Maka tan space open parentheses alpha minus beta close parentheses equals6 6 over 13.

Oleh karena itu, jawaban yang benar adalah A.

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