Menu Close

Question-211251

Tinku Tara 0 Comments
Pin




Question Number 211251 by hardmath last updated on 01/Sep/24
Commented by a.lgnaoui last updated on 04/Sep/24
Commented by a.lgnaoui last updated on 05/Sep/24
((sin C)/(MN))=((sin M)/(BC−AB))=((sin N)/(CD−AD))(1) sin C=((DF)/(FC)) if ∡D=90 then FC=(√(DF^2 +CD^2 )) ((sin C)/(BE))=((sin E)/(BC)) =((AD)/(AE.BC))=((AD)/((BE−AB)BC)) sin C=((BE.AD)/((BE−AB).BC)) BC−AB=a CD−AD=b MN^2 =a^2 +b^2 −2abcos C (i) (1)⇒ ((sin M)/(sin N ))=(a/b) ((sin C)/(sin M))=((MN)/a) ((MN^2 )/a^2 )=((sin^2 C)/(sin^2 M)) (ii) (((i))/b^2 )⇒ ((MN^2 )/b^2 )=[((a/b))^2 +1−2(a/b)cos C (I) (I/((ii)))=(a^2 /b^2 )=(([((a/b))^2 −((2acos C)/b)+1]sin^2 M)/(sin^2 C)) M=∡NMC sin M=(a/b)sin N(N=∡MnC) or MNsin M=BC−AB)sin C and MNcos M+(BC−AB)cos C=CD−AD dinc { ((MNsin M=asin C)),((MNcos M=b−acos C)) :} tan M=((asin C)/(b−acos C)) (1/(cos^2 M))=1+(((a/b)sin^2 C)/((1−(a/b)cos C)^2 )) (1/(sin^2 M))=1+(((1−(a/b)cos C)^2 )/((a/b)sin^2 C)) =(((a/b)sin^2 C+(1−(a/b)cos C)^2 )/((a/b)sin^2 C)) soit ((a/b))=((((a/b)−((2acos C)/b)+1))/([(a/b)sin^2 C+(1−(a/b)cos C)]^2 )) soit x=((x(1−2cos C)+1)/([x(1−cos^2 C)+(1−xcos C)]^2 )) posonz cos C=c c−2xc+1= x^2 (1−c^2 )^2 +(1−xc)^2 +2x(1−c^2 )(1−xc) =x^2 (1+c^4 −/2c^2 )+1+x/c^2 −2/xc+ 2x(1−x/c−c^2 /+x/c^3 ) =(x^2 +2x+1)+x^2 c^2 (c^2 −2)−2x^2 c +2x^2 c^3 −2xc−xc^2 =(x+1)^2 +x^2 [c^2 (c^2 −2)−2c+2c^3 ] −x(c^2 +c) = x^2 (c^4 +2c^3 −2c^2 −2c+1)−x(c^2 −c−2)−c=0 x^2 −(((c^2 −c−2)/(c^4 +2c^3 −2c^2 −2c+1)))x−((c/(c^4 +2c^3 −2c^2 −2c+1)))=0 △=c^2 −c−2)^2 +4c(c^4 +2c^3 −2c^2 −2c+1)=0 c^4 +c^2 +4−2c^3 +4c−4c^2 +4c^5 −8c^4 −8c^2 +4c=0 4c^5 −7c^4 −2c^3 −11c^2 +8c+4=0 c=0,838157... and c<0 alors: x=((−b)/(2a))=((2+c−c^2 )/(2(c^4 +2c^3 −2c^2 −2c+1))) for c=cosC=−0, 338468 soit C=90+56,97° { (( 2+c−c^2 =1,161532)),(( 2(c^4 +2c^3 −2c^2 −2c+1)=1,1706...)) :} alors x=1 ⇒ log_(2024) (((BC−AC)/(CD−AD)))=0 Donc : BC−AC=CD−AD
sinCMN=sinMBC−AB=sinNCD−AD(1)sinC=DFFCif∡D=90thenFC=DF2+CD2sinCBE=sinEBC=ADAE.BC=AD(BE−AB)BCsinC=BE.AD(BE−AB).BCBC−AB=aCD−AD=bMN2=a2+b2−2abcosC(i)(1)⇒sinMsinN=absinCsinM=MNaMN2a2=sin2Csin2M(ii)(i)b2⇒MN2b2=[(ab)2+1−2abcosC(I)I(ii)=a2b2=[(ab)2−2acosCb+1]sin2Msin2CM=∡NMCsinM=absinN(N=∡MnC)orMNsinM=BC−AB)sinCandMNcosM+(BC−AB)cosC=CD−ADdinc{MNsinM=asinCMNcosM=b−acosCtanM=asinCb−acosC1cos2M=1+absin2C(1−abcosC)21sin2M=1+(1−abcosC)2absin2C=absin2C+(1−abcosC)2absin2Csoit(ab)=(ab−2acosCb+1)[absin2C+(1−abcosC)]2soitx=x(1−2cosC)+1[x(1−cos2C)+(1−xcosC)]2posonzcosC=cc−2xc+1=x2(1−c2)2+(1−xc)2+2x(1−c2)(1−xc)=x2(1+c4−/2c2)+1+x/c2−2/xc+2x(1−x/c−c2/+x/c3)=(x2+2x+1)+x2c2(c2−2)−2x2c+2x2c3−2xc−xc2=(x+1)2+x2[c2(c2−2)−2c+2c3]−x(c2+c)=x2(c4+2c3−2c2−2c+1)−x(c2−c−2)−c=0x2−(c2−c−2c4+2c3−2c2−2c+1)x−(cc4+2c3−2c2−2c+1)=0△=c2−c−2)2+4c(c4+2c3−2c2−2c+1)=0c4+c2+4−2c3+4c−4c2+4c5−8c4−8c2+4c=04c5−7c4−2c3−11c2+8c+4=0c=0,838157…andc<0alors:x=−b2a=2+c−c22(c4+2c3−2c2−2c+1)forc=cosC=−0,338468soitC=90+56,97°{2+c−c2=1,1615322(c4+2c3−2c2−2c+1)=1,1706…alorsx=1⇒log2024(BC−ACCD−AD)=0Donc:BC−AC=CD−AD

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *