Pertanyaan

Sebuah △ PQR dengan ∠ PQR = 4 5 ∘ , ∠ SPR = 3 0 ∘ ,dan QR = 6 cm . Tentukan PQ dan PR ! Gambar 6.23 Segitiga

Sebuah $△ PQR$ dengan $∠ PQR = 4 5^{\circ}$ , $∠ SPR = 3 0^{\circ}$ , dan $QR = 6 cm$ . Tentukan $PQ$ dan $PR$ !

Gambar 6.23 Segitiga

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Klaim

E. Nur

Master Teacher

Mahasiswa/Alumni Institut Teknologi Sepuluh Nopember

Jawaban terverifikasi

Jawaban

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Pembahasan

Perhatikan gambar berikut. 105 Gunakan aturan sinus. Jadi, Jadi, .

Perhatikan gambar berikut.

105

Gunakan aturan sinus.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator PQ over denominator sin open parentheses 105 degree close parentheses end fraction end cell equals cell fraction numerator QR over denominator sin open parentheses 30 degree close parentheses end fraction end cell row cell fraction numerator PQ over denominator sin open parentheses 60 degree plus 45 degree close parentheses end fraction end cell equals cell fraction numerator QR over denominator sin open parentheses 30 degree close parentheses end fraction end cell row cell fraction numerator PQ over denominator sin open parentheses 60 degree close parentheses cos open parentheses 45 degree close parentheses plus cos open parentheses 60 degree close parentheses sin open parentheses 45 degree close parentheses end fraction end cell equals cell fraction numerator 6 over denominator begin display style 1 half end style end fraction end cell row cell fraction numerator PQ over denominator begin display style 1 half end style square root of 3 times begin display style 1 half end style square root of 2 plus begin display style 1 half end style times begin display style 1 half end style square root of 2 end fraction end cell equals cell 12 over 1 end cell row cell fraction numerator PQ over denominator begin display style 1 fourth square root of 6 plus 1 fourth square root of 2 end style end fraction end cell equals cell 12 over 1 end cell row PQ equals cell 12 cross times 1 fourth square root of 2 open parentheses square root of 3 plus 1 close parentheses end cell row PQ equals cell 3 square root of 2 open parentheses square root of 3 plus 1 close parentheses end cell end table end style$

Jadi,

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator PR over denominator sin open parentheses 45 degree close parentheses end fraction end cell equals cell fraction numerator QR over denominator sin open parentheses 30 degree close parentheses end fraction end cell row cell fraction numerator PR over denominator begin display style 1 half end style square root of 2 end fraction end cell equals cell fraction numerator 6 over denominator begin display style 1 half end style end fraction end cell row PR equals cell 6 square root of 2 end cell end table end style$

Jadi, .