VA rogg 40 de puncte...Dau coroana
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Răspuns:
Explicație pas cu pas:
AB=8cm, BC=6cm, C'C=10cm,
a) m(A'C, (ABC))=???
A'C oblica pe (ABC), A'A⊥(ABC), AC=pr(ABC)A'C. Atunci m(A'C, (ABC))=m(∡A'C,AC)=m(∡A'CA).
ABCD dreptunghi, deci AC²=AB²+BC²=8²+6²=10², deci AC=10cm
In ΔA'AC, dreptunghic in A, A'A=AC=10, deci In ΔA'AC dreptunghic isoscel, atunci m(∡A'CA)=45°=m(A'C, (ABC)).
b) sin(∡(D'B, (ADA')))=???
BD' oblica la (ADA'), BA⊥(ADA'), deci AD'=pr(ADA')BD'. Atunci
∡(D'B, (ADA'))=∡(D'B,AD')=∡BD'A. ΔBAD' dreptunghic in A.
sin(∡BD'A)=AB/BD', dar BD'²=BD²+DD'²=10²+10²=10²·2, deci BD'=10√2
Atunci sin(∡BD'A)=AB/BD'=8/(10√2)=8√2/(10·2)=2√2/5= sin(∡(D'B, (ADA'))).
c) tg(∡(BD',(DCC')))=???
BD' oblica la (DCC'), BC⊥(DCC'), D'C=pr(DCC')BD', atunci ∡(BD',(DCC'))=∡(BD',D'C)=∡BD'C. ΔBD'C dreptunghic in C, deci tg(∡BD'C)=BC/D'C, BC=6, D'C²=D'B²-BC²=(10√2)²-6²=200-36=164=4·41
Deci D'C=2√41cm.
Atunci tg(∡BD'C)=BC/D'C=6/(2√41)=3√41/41= tg(∡(BD',(DCC'))).