Diketahui dan berada dikuadran IV. Nilai adalah ....

Pertanyaan

Diketahui $cosec space straight beta equals negative 13 over 12$ dan $straight beta$ berada dikuadran IV. Nilai $tan space straight beta$ adalah ....

$negative 13 over 5$
$negative 12 over 5$
$negative 12 over 13$
$negative 5 over 12$
$negative 5 over 13$

Pembahasan Soal:

Diketahui:

$cosec space straight beta equals fraction numerator 1 over denominator sin space straight beta end fraction equals negative 13 over 12 sin space straight beta equals negative 12 over 13$

Misalkan: $sin space straight beta equals y over r$

$table attributes columnalign right center left columnspacing 0px end attributes row x equals cell square root of r squared minus y squared end root end cell row blank equals cell square root of 13 squared minus 12 squared end root end cell row blank equals cell square root of 169 minus 144 end root end cell row blank equals cell square root of 25 end cell row blank equals 5 end table$

Maka, nilai $tan space straight beta equals negative straight y over straight x equals negative 12 over 5$

Catatan: nilai $sin space straight beta$ dan $tan space straight beta$ di kuadran IV adalah negatif

Jadi, jawaban yang tepat adalah B.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Nur

Mahasiswa/Alumni Universitas Muhammadiyah Malang

Terakhir diupdate 05 Juni 2021

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Pembahasan Soal:

Dari soal tersebut, pertama-tama tentukan panjang sisi miring (OT).

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row s equals cell square root of x squared plus y squared end root end cell row blank equals cell square root of open parentheses negative 3 close parentheses squared plus 2 squared end root end cell row blank equals cell square root of 13 end cell end table end style$

dengan begitu, dapat diperoleh nilai-nilai trigonemtri sebagai berikut:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell y over s end cell row blank equals cell fraction numerator 2 over denominator square root of 13 end fraction end cell row blank equals cell 2 over 13 square root of 13 end cell row cell cos space alpha end cell equals cell x over s end cell row blank equals cell fraction numerator negative 3 over denominator square root of 13 end fraction end cell row blank equals cell negative 3 over 13 square root of 13 end cell row cell tan space alpha end cell equals cell y over x end cell row blank equals cell fraction numerator 2 over denominator negative 3 end fraction end cell row blank equals cell negative 2 over 3 end cell end table end style$ $begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell s e c space alpha end cell equals cell fraction numerator 1 over denominator cos space alpha end fraction end cell row blank equals cell space minus 1 third square root of 13 end cell row cell cos e c space alpha end cell equals cell fraction numerator 1 over denominator sin space alpha end fraction end cell row blank equals cell 1 half square root of 13 end cell row cell c o tan space alpha end cell equals cell fraction numerator 1 over denominator tan space alpha end fraction end cell row blank equals cell negative 3 over 2 end cell end table end style$

Roboguru

Pada uraian materi di atas diperoleh pada segitiga siku-siku (sudut di kuadran I) berlaku identitas trigonometri berikut. 1. Buktikan kedelapan identitas trigonometri tersebut berlaku untuk sudut ...

Pembahasan Soal:

Pada kuadran II :

$begin mathsize 14px style sin space alpha equals fraction numerator 1 over denominator cosec space alpha end fraction greater than 0 end style$

Pada kuadran III dan IV:

$begin mathsize 14px style sin space alpha equals fraction numerator 1 over denominator cosec space alpha end fraction space less than 0 end style$ .

Roboguru

Lengkapilah tabel berikut!

Pembahasan Soal:

Karena $sin space 37 degree equals 0 comma 6$ , maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space 37 degree end cell equals cell 6 over 10 end cell row blank equals cell 3 over 5 end cell row cell de over mi end cell equals cell 3 over 5 end cell end table$

Dengan menggambar segitiga siku-siku yang bersesuaian, asumsikan panjang sisi depan sudut = 3 satuan dan panjang sisi miring = 5 satuan, sebagai berikut:

Dari segitiga diatas, dapat ditentukan sisi samping sudut, yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row sa equals cell square root of mi squared minus de squared end root end cell row blank equals cell square root of 5 squared minus 3 squared end root end cell row blank equals cell square root of 25 minus 9 end root end cell row blank equals cell square root of 16 end cell row blank equals 4 end table$

Sehingga, didapat:

$table attributes columnalign right center left columnspacing 0px end attributes row cell cos space 37 degree end cell equals cell sa over mi end cell row blank equals cell 4 over 5 end cell row blank equals cell 0 comma 8 end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row cell tan space 37 degree end cell equals cell de over sa end cell row blank equals cell 3 over 4 end cell row blank equals cell 0 comma 75 end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row cell cosec space 37 degree end cell equals cell fraction numerator 1 over denominator sin space 37 degree end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style 3 over 5 end style end fraction end cell row blank equals cell 5 over 3 end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row cell sec space 37 degree end cell equals cell fraction numerator 1 over denominator cos space 37 degree end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style 4 over 5 end style end fraction end cell row blank equals cell 5 over 4 end cell row blank equals cell 1 comma 25 end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row cell cot space 37 degree end cell equals cell fraction numerator 1 over denominator tan space 37 degree end fraction end cell row blank equals cell fraction numerator 1 over denominator begin display style 3 over 4 end style end fraction end cell row blank equals cell 4 over 3 end cell end table$

Jadi, tabel yang telah dilengkapi adalah sebagai berikut.

Roboguru

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Pembahasan Soal:

Kita cari sisi miring nya

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perbandingan trigonometri sudut $begin mathsize 14px style straight alpha end style$

$begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell sin space straight alpha end cell equals cell de over mi equals 6 over 10 end cell row cell cos space straight alpha end cell equals cell sa over mi equals 8 over 10 end cell row cell tan space straight alpha end cell equals cell de over sa equals 6 over 8 end cell row cell cossec space straight alpha end cell equals cell fraction numerator 1 over denominator sin space straight alpha end fraction equals 10 over 6 end cell row cell sec space straight alpha end cell equals cell fraction numerator 1 over denominator cos space straight alpha end fraction equals 10 over 8 end cell row cell cot space straight alpha end cell equals cell fraction numerator 1 over denominator tan space straight alpha end fraction equals 8 over 6 end cell end table end style$

Roboguru

Tentukan nilai dari keenam perbandingan trigonometri sudut pada segitiga siku-siku berikut.

Pembahasan Soal:

Pertama kita cari sisi a

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$begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell sin space straight alpha end cell equals cell de over mi equals 24 over 25 end cell row cell cos space straight alpha end cell equals cell sa over mi equals 7 over 25 end cell row cell tan space straight alpha end cell equals cell de over sa equals 24 over 7 end cell row cell cossec space straight alpha end cell equals cell fraction numerator 1 over denominator sin space straight alpha end fraction equals 25 over 24 end cell row cell sec space straight alpha end cell equals cell fraction numerator 1 over denominator cos space straight alpha end fraction equals 25 over 7 end cell row cell cot space straight alpha end cell equals cell fraction numerator 1 over denominator tan space straight alpha end fraction equals 7 over 24 end cell end table end style$

Roboguru