Pertanyaan

Diketahui kubus ABCD.EFGH. titik M berada di rusuk AD sedemikian sehingga AM : MD = 1 : 2. Titik N berada di rusuk CD sedemikian sehingga CN : ND = 1 : 2. Titik P berada di rusuk DH sedemikian sehingga DP : PH = 2 : 1. Jika α adalah sudut antara bidang MNP dan garis PB maka nilai cos ⁡α = ….

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A. Rizky

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

begin mathsize 14px style bold I bold. bold space bold ILUSTRASI bold space bold GAMBAR bold space bold KUBUS bold space bold SEPERTI bold space bold SOAL bold I bold. bold space bold Identifikasi bold space bold sisi bold minus bold sisi bold space bold kubusnya end style

$begin mathsize 14px style Perhatikan space segitiga space BPQ comma space sehingga space kita space dapat space menentukan space nilai space cos space straight x space nya comma yaitu space menggunakan space aturan space cosinus space colon Cos space straight x space equals space fraction numerator BP squared plus PQ squared minus BQ squared over denominator 2 straight space BP. PQ end fraction equals straight space fraction numerator open parentheses straight a square root of 22 straight space end root close parentheses squared plus left parenthesis straight a square root of 6 straight space end root right parenthesis squared minus open parentheses 2 straight a square root of 2 straight space end root straight space close parentheses squared over denominator 2. straight space open parentheses straight a square root of 22 straight space end root close parentheses straight space. left parenthesis straight a square root of 6 straight space end root right parenthesis end fraction equals fraction numerator 22 straight a squared plus 6 straight a squared minus 8 straight a squared over denominator 2 straight a squared square root of 132 straight space end root straight space end fraction space space space space space space space space space space space equals space fraction numerator 10 over denominator 2 square root of 33 straight space end root straight space end fraction equals fraction numerator 5 over denominator square root of 33 straight space end root straight space end fraction left parenthesis pembilang space dan space penyebut space dikalikan space √ left parenthesis 33 space right parenthesis right parenthesis comma space maka space hasilnya space space space space space space space space space space equals fraction numerator 5 square root of 33 straight space end root over denominator 33 straight space end fraction end style$

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