Jika dan . Tentukan semua perbandingan trigonometri sudut dan sudut .

Pertanyaan

Jika $angle ACB equals alpha$ dan $angle ECD equals beta$ . Tentukan semua perbandingan trigonometri sudut $alpha$ dan sudut $beta$ .

$space$

Pembahasan Soal:

Ingat!

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell fraction numerator sisi space depan space sudut space alpha over denominator sisi space miring end fraction end cell row cell cos space alpha end cell equals cell fraction numerator sisi space samping space sudut space alpha over denominator sisi space miring end fraction end cell row cell tan space alpha end cell equals cell fraction numerator sisi space depan space sudut space alpha over denominator sisi space samping space sudut space alpha end fraction end cell row cell cosec space alpha end cell equals cell fraction numerator 1 over denominator sin space alpha end fraction end cell row cell sec space alpha end cell equals cell fraction numerator 1 over denominator cos space alpha end fraction end cell row cell cot space alpha end cell equals cell fraction numerator 1 over denominator tan space alpha end fraction end cell end table$

Berdasarkan gambar, diketahui:

$table attributes columnalign right center left columnspacing 0px end attributes row cell angle ACB end cell equals cell angle DCE space open parentheses bertolak space belakang close parentheses end cell row cell angle CAB end cell equals cell angle CDE equals 90 degree end cell end table$

Maka, $angle ABC equals angle DEC$ .

Sehingga, segitiga CAB sebangun dengan segitiga CDE.

Akibatnya perbandingan sisi-sisi kedua segitiga tersebut sama besar.

Tentukan panjang DE.

$table attributes columnalign right center left columnspacing 0px end attributes row cell DE over AB end cell equals cell CD over CA end cell row cell DE over 6 end cell equals cell 24 over 8 end cell row DE equals cell 3 times 6 end cell row DE equals cell 18 space cm end cell end table$

Karena segitiga CAB dan segitiga CDE merupakan segitiga siku-siku, maka gunakan teorema Pythagoras untuk menentukan panjang BC dan CE.

$table attributes columnalign right center left columnspacing 0px end attributes row BC equals cell square root of AC squared plus BC squared end root end cell row blank equals cell square root of 8 squared plus 6 squared end root end cell row blank equals cell square root of 100 end cell row blank equals cell 10 space cm end cell row CE equals cell square root of CD squared plus DE squared end root end cell row blank equals cell square root of 24 squared plus 18 squared end root end cell row blank equals cell square root of 900 end cell row blank equals cell 30 space cm end cell end table$

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank angle end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank ACB end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank angle end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank DCE end table$ , maka $alpha equals beta$ . Akibatnya nilai perbandingan trigonometri sudut $alpha$ sama dengan nilai perbandingan trigonometri sudut $beta$ .

Nilai perbandingan trigonimetri sudut $alpha$ .

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell AB over CB end cell row cell sin space alpha end cell equals cell 6 over 10 end cell row cell sin space alpha end cell equals cell 3 over 5 end cell row cell cos space alpha end cell equals cell AC over BC end cell row cell cos space alpha end cell equals cell 8 over 10 end cell row cell cos space alpha end cell equals cell 4 over 5 end cell row cell tan space alpha end cell equals cell AB over AC end cell row cell tan space alpha end cell equals cell 6 over 8 end cell row cell tan space alpha end cell equals cell 3 over 4 end cell row cell cosec space alpha end cell equals cell fraction numerator 1 over denominator begin display style bevelled 3 over 5 end style end fraction end cell row cell cosec space alpha end cell equals cell 5 over 3 end cell row cell sec space alpha end cell equals cell fraction numerator 1 over denominator begin display style bevelled 4 over 5 end style end fraction end cell row cell sec space alpha end cell equals cell 5 over 4 end cell row cell cot space alpha end cell equals cell fraction numerator 1 over denominator begin display style bevelled 3 over 4 end style end fraction end cell row cell cot space alpha end cell equals cell 4 over 3 end cell end table$

Jadi,

$table attributes columnalign right center left columnspacing 0px end attributes row cell sin space alpha end cell equals cell sin space beta equals 3 over 5 end cell row cell cosec space alpha end cell equals cell cosec space beta equals 5 over 3 end cell row cell cos space alpha end cell equals cell cos space beta equals 4 over 5 end cell row cell sec space alpha end cell equals cell sec space beta equals 5 over 4 end cell row cell tan space alpha end cell equals cell tan space beta equals 3 over 4 end cell row cell cot space alpha end cell equals cell cot space beta equals 4 over 3 end cell end table$

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 03 Juni 2021

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