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Jawab1) Kubus ABCD, EFGH,
misal rusuk = 2 cm maka diagonal sisi = 2√2
T titik potong AG dan FH
GT = 1/2 EG = 1/2 (2√2) = √2
Hubungkan C dengan T = CT
> (GC, dengan bidang AFH) = < GCT = α
ΔCGT siku siku di G
Tan α = GT/CG = √2/(2) = 1/2 √2
2)
Kubus ABCD, EFGH
misalkan rusuk = 2 cm
diagonal sisi = 2√2 cm
T titik ptong EG dan FH
ET = 1/2 EG = 1/2 (2√2) = √2
AET siku siku di E
AT = √(AE² + ET²) = √(2² + (√2)²) = √6
Pada Δ ATC
AT = TC = √6
AC = 2√2
< TC dengan bidang AFH =α= < (CT, TA) = <ATC
cos <ATC = (AT² + CT²- AC²) / (2 . AT. CT))
cos α = (6 + 6 - 8) / (2.√6. √6)
cos α = 4/(12) = 1/3
sin α = √(1 - cos² α)
sin α = √( 1 - (1/3)² = √(1 - 1/9) = √(8/9)
sin α = ²/₃ √2