Pertanyaan

Untuk menentukan tinggi bukit ( t ), seorang pengamat mengukur sudut elevasi ( α dan β ) dari titik A dan B yang berjarak d . Tunjukkan: b. t = sin ( α − β ) d sin α ⋅ cos β

Untuk menentukan tinggi bukit ( $t$ ), seorang pengamat mengukur sudut elevasi ( $α$ dan $β$ ) dari titik A dan B yang berjarak $d$ .

Tunjukkan:

b. $t = \frac{d sin α \cdot cos β}{sin ( α - β )}$ $\;$

A. Armanda

Master Teacher

Mahasiswa/Alumni Universitas Indraprasta PGRI

Jawaban terverifikasi

Jawaban

terbukti bahwa .

terbukti bahwa $table attributes columnalign right center left columnspacing 0px end attributes row blank blank t end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator d space sin space alpha space sin space beta over denominator sin space open parentheses alpha space minus beta close parentheses end fraction end cell end table$ .

Pembahasan

Nilai tangen diperoleh dari perbandingan sisi depan sudut dengan sisi samping sudut pada segitiga siku-siku. Ingat! Selisih Dua Sudut pada Sinus Misalkan sisi samping adalah . Berdasarkan gambar di atas, diketahui: Maka, diperoleh: Terdapat kesalahan pada soal . Bukan , melainkan . Akan ditunjukkan . Substitusikan persamaan (i) dan (ii). Jadi, terbukti bahwa .

Nilai tangen diperoleh dari perbandingan sisi depan sudut dengan sisi samping sudut pada segitiga siku-siku.

Ingat!

$tan space alpha equals fraction numerator sin space alpha over denominator cos space alpha end fraction$

Selisih Dua Sudut pada Sinus

sin space open parentheses alpha minus beta close parentheses equals sin space alpha space cos space beta minus cos space alpha space sin space beta

Misalkan sisi samping alpha adalah .

Berdasarkan gambar di atas, diketahui:

$table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell t over x end cell row cell tan space beta end cell equals cell fraction numerator t over denominator d plus x end fraction end cell end table$

Maka, diperoleh:

$table attributes columnalign right center left columnspacing 0px end attributes row cell tan space alpha end cell equals cell t over x end cell row x equals cell fraction numerator t over denominator tan space alpha end fraction space... space left parenthesis straight i right parenthesis end cell row blank blank blank row cell tan space beta end cell equals cell fraction numerator t over denominator d plus x end fraction end cell row cell d plus x end cell equals cell fraction numerator t over denominator tan space beta end fraction end cell row x equals cell fraction numerator t over denominator tan space beta end fraction minus d space... space left parenthesis ii right parenthesis end cell end table$

Terdapat kesalahan pada soal. Bukan $t equals fraction numerator d space sin space alpha times cos space beta over denominator sin space open parentheses alpha minus beta close parentheses end fraction$ , melainkan $table attributes columnalign right center left columnspacing 0px end attributes row blank blank t end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator d space sin space alpha space sin space beta over denominator sin space open parentheses alpha space minus beta close parentheses end fraction end cell end table$ .

Akan ditunjukkan $table attributes columnalign right center left columnspacing 0px end attributes row blank blank t end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator d space sin space alpha space sin space beta over denominator sin space open parentheses alpha space minus beta close parentheses end fraction end cell end table$ .

Substitusikan persamaan (i) dan (ii).

$table attributes columnalign right center left columnspacing 0px end attributes row x equals x row cell fraction numerator t over denominator tan space alpha end fraction end cell equals cell fraction numerator t over denominator tan space beta end fraction minus d end cell row cell fraction numerator t over denominator tan space alpha end fraction times tan space alpha space tan space beta end cell equals cell open parentheses fraction numerator t over denominator tan space beta end fraction minus d close parentheses times tan space alpha space tan space beta end cell row cell t space tan space beta end cell equals cell t space tan space alpha minus d space tan space alpha space tan space beta end cell row cell d space tan space alpha space tan space beta end cell equals cell t space tan space alpha minus t space tan space beta end cell row cell d space tan space alpha space tan space beta end cell equals cell t space open parentheses tan space alpha minus tan space beta close parentheses end cell row cell fraction numerator d space tan space alpha space tan space beta over denominator tan space alpha minus tan space beta end fraction end cell equals t row t equals cell fraction numerator d space tan space alpha space tan space beta over denominator tan space alpha minus tan space beta end fraction end cell row t equals cell fraction numerator begin display style fraction numerator d space sin space alpha space sin space beta over denominator cos space alpha space cos space beta end fraction end style over denominator begin display style fraction numerator sin space alpha over denominator cos space alpha end fraction end style minus fraction numerator sin space beta over denominator cos space beta end fraction end fraction end cell row t equals cell fraction numerator begin display style fraction numerator d space sin space alpha space sin space beta over denominator up diagonal strike cos space alpha space cos space beta end strike end fraction end style over denominator begin display style fraction numerator sin space alpha space cos space beta minus cos space alpha space sin space beta over denominator up diagonal strike cos space alpha space cos space beta end strike end fraction end style end fraction end cell row t equals cell fraction numerator d space sin space alpha space sin space beta over denominator sin space alpha space cos space beta minus cos space alpha space sin space beta end fraction end cell row t equals cell fraction numerator d space sin space alpha space sin space beta over denominator sin space open parentheses alpha space minus beta close parentheses end fraction end cell end table$

Jadi, terbukti bahwa $table attributes columnalign right center left columnspacing 0px end attributes row blank blank t end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator d space sin space alpha space sin space beta over denominator sin space open parentheses alpha space minus beta close parentheses end fraction end cell end table$ .