Diketah kubus ABCD.EFGH dengan rusuk 4 cm. tangen sudut antara garis BF dan bidang BEG adalah.... A.1/2√6 B.1/2√3 C.1/2√2
rirismaulidina
O = tengah FH BO = √(2√2)² + 4² = √24 FO ² = BO² + BF² - 2 x BO x BF cos α (2√2)² = (√24)² + 4² - 2√24x4 cosα 8 = 24 +16 -8 √24 cosα cos α = √6/3 √6 = x 3 = r y = r² - x² = √3 tan = y / x tan = √3 per √6 = 1/2 √6 ( A )
BO = √(2√2)² + 4² = √24
FO ² = BO² + BF² - 2 x BO x BF cos α
(2√2)² = (√24)² + 4² - 2√24x4 cosα
8 = 24 +16 -8 √24 cosα
cos α = √6/3
√6 = x
3 = r
y = r² - x² = √3
tan = y / x
tan = √3 per √6 = 1/2 √6 ( A )