Pertanyaan

Segitiga ABC siku-siku di C. Jika sin A = k , maka tan ( A − B ) = ...

Segitiga ABC siku-siku di C. Jika $sin A = k$ , maka $tan (A - B) =$ ...

$fraction numerator straight k squared minus 1 over denominator 2 straight k square root of 1 minus straight k squared end root end fraction$
$fraction numerator 1 minus straight k squared over denominator 2 straight k square root of 1 minus straight k squared end root end fraction$
$fraction numerator 2 straight k squared minus 1 over denominator 2 straight k square root of 1 minus straight k squared end root end fraction$
$fraction numerator 1 minus 2 straight k squared over denominator 2 straight k square root of 1 minus straight k squared end root end fraction$
$fraction numerator straight k squared minus 1 over denominator straight k square root of 1 minus straight k squared end root end fraction$

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Habis dalam

Klaim

R. Hajrianti

Master Teacher

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah C.

Pembahasan

Diketahui: Segitiga ABC siku-siku di Cdengan . Sehingga diperoleh: Dengan menggunakan Teorema Pythagoras, diperoleh panjang sisi samping sudut A yaitu: Soal digambarkan sebagai berikut: Sehingga diperoleh: Rumus tangen untuk selisih dua sudut yaitu: Diperoleh penyelesaiannya yaitu: Maka . Oleh karena itu, jawaban yang benar adalah C.

Diketahui: Segitiga ABC siku-siku di Cdengan sin space straight A equals straight k .

Sehingga diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space straight A end cell equals straight k row cell fraction numerator depan space angle straight A over denominator miring space angle straight A end fraction end cell equals cell straight k over 1 end cell end table$

Dengan menggunakan Teorema Pythagoras, diperoleh panjang sisi samping sudut A yaitu:

Soal digambarkan sebagai berikut:

Sehingga diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell tan space straight A end cell equals cell fraction numerator depan space angle straight A over denominator samping space angle straight A end fraction end cell row blank equals cell fraction numerator straight k over denominator square root of 1 minus straight k squared end root end fraction end cell row blank blank blank row cell tan space straight B end cell equals cell fraction numerator depan space angle straight B over denominator samping space angle straight B end fraction end cell row blank equals cell fraction numerator square root of 1 minus straight k squared end root over denominator straight k end fraction end cell end table$

Rumus tangen untuk 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 straight A minus straight B close parentheses end cell equals cell fraction numerator tan space straight A minus tan space straight B over denominator 1 plus tan space straight A times tan space straight B end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator straight k over denominator square root of 1 minus straight k squared end root end fraction end style minus begin display style fraction numerator square root of 1 minus straight k squared end root over denominator straight k end fraction end style over denominator 1 plus begin display style fraction numerator down diagonal strike straight k over denominator down diagonal strike square root of 1 minus straight k squared end root end strike end fraction end style times begin display style fraction numerator down diagonal strike square root of 1 minus straight k squared end root end strike over denominator down diagonal strike straight k end fraction end style end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator straight k over denominator square root of 1 minus straight k squared end root end fraction minus fraction numerator square root of 1 minus straight k squared end root over denominator straight k end fraction end style over denominator 1 plus 1 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator open parentheses straight k close parentheses open parentheses straight k close parentheses over denominator open parentheses square root of 1 minus straight k squared end root close parentheses open parentheses straight k close parentheses end fraction minus fraction numerator open parentheses square root of 1 minus straight k squared end root close parentheses open parentheses square root of 1 minus straight k squared end root close parentheses over denominator open parentheses straight k close parentheses open parentheses square root of 1 minus straight k squared end root close parentheses end fraction end style over denominator 2 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator straight k squared over denominator open parentheses square root of 1 minus straight k squared end root close parentheses open parentheses straight k close parentheses end fraction minus fraction numerator 1 minus straight k squared over denominator open parentheses straight k close parentheses open parentheses square root of 1 minus straight k squared end root close parentheses end fraction end style over denominator 2 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator straight k squared minus 1 plus straight k squared over denominator straight k square root of 1 minus straight k squared end root end fraction end style over denominator 2 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 2 straight k squared minus 1 over denominator straight k square root of 1 minus straight k squared end root end fraction end style over denominator 2 end fraction end cell row blank equals cell fraction numerator 2 straight k squared minus 1 over denominator straight k square root of 1 minus straight k squared end root end fraction times 1 half end cell row blank equals cell fraction numerator 2 straight k squared minus 1 over denominator 2 straight k square root of 1 minus straight k squared end root end fraction end cell end table$

Maka tan space open parentheses straight A minus straight B close parentheses equals $fraction numerator 2 straight k squared minus 1 over denominator 2 straight k square root of 1 minus straight k squared end root end fraction$ .

Oleh karena itu, jawaban yang benar adalah C.