Diketahui kubus ABCD.EFGH dengan panjang rusuk "a" cm. nilai kosinus sudut antara bidang ABCD dengan bidang BDG adalah.... a. akar dua b. sepertiga akar tiga c. setengah akar tiga d sepertiga akar enam e. setengah akar enam
r = a cm perhatikan segitiga PCG! CG = r = a cm AC = diagonal sisi = a√2 PC = 1/2 AC = 1/2 a √2 cm PG dicari dgn Pythagoras: PG² = CG² + PC² PG² = a² + (1/2a√2)² PG² = a² + 1/2a² PG² = 3/2a² PG = √(3/2a²) PG = 1/2 a √6
nilai kosinus = sisi samping / sisi miring cos P = PC/PG = (1/2 a √2) / (1/2 a √6) = √2 / √6 x √6 / √6 = 1/6 √12 = 1/6 x 2√3 = 2/6 √3 = 1/3 √3
r = a cm
perhatikan segitiga PCG!
CG = r = a cm
AC = diagonal sisi = a√2
PC = 1/2 AC = 1/2 a √2 cm
PG dicari dgn Pythagoras:
PG² = CG² + PC²
PG² = a² + (1/2a√2)²
PG² = a² + 1/2a²
PG² = 3/2a²
PG = √(3/2a²)
PG = 1/2 a √6
nilai kosinus = sisi samping / sisi miring
cos P = PC/PG
= (1/2 a √2) / (1/2 a √6)
= √2 / √6 x √6 / √6
= 1/6 √12
= 1/6 x 2√3
= 2/6 √3
= 1/3 √3
semoga membantu ya :)