PEMBAHASAN
Dimensi Tiga
Kubus
r = 4 cm
O tengah EG
Perhatikan ∆OFB siku di F
Sudut antara BF dan bidang BEG adalah sudut FBO.
∠(BF , BEG) = ∠FBO = ∠B
diagonal bidang FH = r√2
FO = 1/2 FH = 1/2 r√2
tan B
= FO/FB
= (1/2 r√2)/r
= 1/2 √2
∠B = arctan (1/2 √2)
∠B ≈ 35,26°
•
BO = 1/2 r√6
sin B
= FO/BO
= (1/2 r√2) / (1/2 r√6)
= 1/√3
= 1/3 √3
cos B
= BF/BO
= r / (1/2 r√6)
= 2/√6
= 1/3 √6
••
Asal rumus cepat dr BO
BO
= √(FB² + FO²)
= √(r² + (1/2 r√2)²)
= √(r² + 2/4 r²)
= √(r²/4 (4 + 2))
= 1/2 r √6
= 1/2 × 4√6
= 2√6 cm
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PEMBAHASAN
Dimensi Tiga
Kubus
r = 4 cm
O tengah EG
Perhatikan ∆OFB siku di F
Sudut antara BF dan bidang BEG adalah sudut FBO.
∠(BF , BEG) = ∠FBO = ∠B
diagonal bidang FH = r√2
FO = 1/2 FH = 1/2 r√2
tan B
= FO/FB
= (1/2 r√2)/r
= 1/2 √2
∠B = arctan (1/2 √2)
∠B ≈ 35,26°
•
BO = 1/2 r√6
sin B
= FO/BO
= (1/2 r√2) / (1/2 r√6)
= 1/√3
= 1/3 √3
cos B
= BF/BO
= r / (1/2 r√6)
= 2/√6
= 1/3 √6
••
Asal rumus cepat dr BO
BO
= √(FB² + FO²)
= √(r² + (1/2 r√2)²)
= √(r² + 2/4 r²)
= √(r²/4 (4 + 2))
= 1/2 r √6
= 1/2 × 4√6
= 2√6 cm