1/97 de jos
din punctul P se coboara perpendicular pe MN , la intersectie rezulta punctul R
MQ=3PQ=6 CM, MQ= 6 CM SI PQ=6:3=2 cm
in triunghiul RPN dreptunghic : cu unghiul N =45°
sin 45=PR/PN dar PR =QM = 6 cm
PN= PR / (√2/2)
PN=(6×2)/√2=6√2 (dupa rationalizare)
MN = MR +RN (MR=QP= 2cm)
RN in triunghiul RPN
cos 45°=RN/PN deci √2/2=RN/6√2
RN=6√2×(√2/2)
RN = 6cm
MN = MR +RN (=2+6=8 cm
deci Perimetrul= QP+PN+NM+QM=2+6√2+8+6=16+6√2=2(8+3√2) cm
2 de sus
M(∡MNP)=90°; sin P=0,75 ; p=16; NQ⊥MP, Q∈MP
sin P=MN/MP
MN=16×0,75
MN=12 cm
NP²=MP²-MN²
=256-144
NP=4√7 cm
3. de sus
ctg E=2; ED =4
cotangenta=cateta alaturata / cateta opusa
ctg E=ED/DF
2 = 4/DF⇒ DF =2 cm
FE²= DE²+DF²=
FE =√(4²+2²)
FE = 2√5 cm
4 de sus
C=30°
AD⊥BC
D∈BC
BC=20
sin 30°=AB/BC 1/2=AB/BC
AB=BC/2=20/2=10 cm
AC²= BC²-AB²=
AC=√400-100
AC=10√3 cm
5. de sus
AC=6√2
AC=4√3
BC=2√30
sin =cateta opusa/ipotenuza; cos=cateta alaturata/ipotenuza
ctg=cateta alaturata/cateta opusa si tg=catetea opusa/ cateta alaturata
verificam daca triunghiul este dreptunghic
BC²=AC²+AB²
(2√30)²=(4√3)²+(6√3)²
(√120)²=(√3×16)²+(√2×6²)²
120=48+72
120=120 deci triunghiul este dreptunghic in A
AB=6√2 si AC =4√3 si BC =2√30
sin B=AC/BC=(4√3)/(2√30)=√10/5
cos B=AB/BC=(6√2)/(2√30)=
tg C=AB/AC=(6√2)/(2√30)=
ctg C= AC/AB=(4√3)/(2√30)=
