Răspuns:
Explicație pas cu pas:
a) BD⊥α, D∈α, ΔABD dreptunghic in D. deci d(B,α)=BD. atunci m(∡(AB,α))=m(∡(AB,AD))=45°.
sin(∡BAD)=BD/AB, ⇒BD/9=√2/2, deci BD=9√2/2 cm = d(B,α).
b) ∡(BC,α)=∡(BC,CD)=∡BCD, Daca BD⊥α, atunci BD⊥CD, deci ΔBCD dreptunghic in D.
sin(∡BCD)=BD/BC.
Aflam BC din ΔABC, dreptunghic in B. BC²=AC²-AB²=15²-9²=3²·5²-3²·3²=3²·(5²-3²)=3²·16, deci BC=√(3²·16)=3·4=12.
Atunci sin(∡BCD)=BD/BC=9√2/2 : 12=3√2/2 :4=3√2/8=sin(∠(BC,α))
c) BM⊥AC, ⇒DM⊥AC. Din formula ariei ΔABC, avem:
AC·BM=AB·BC, inlocuim, 15·BM=9·12, ⇒BM=(9·12)/15=36/5
Atunci sin(∡(BM,α))=BD/BM=9√2/2: 36/5=9√2/2 · 5/36=5√2/8.
