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Question-80237




Question Number 80237 by M±th+et£s last updated on 01/Feb/20
Commented by M±th+et£s last updated on 01/Feb/20
if ABCDEFG is a heptagon and AB=1  HIJKLM is hexagon   than find HI
ifABCDEFGisaheptagonandAB=1HIJKLMishexagonthanfindHI
Commented by mr W last updated on 01/Feb/20
HIJKLMN is also a heptagon!
HIJKLMNisalsoaheptagon!
Commented by M±th+et£s last updated on 01/Feb/20
you are right sir its typo
youarerightsiritstypo
Commented by Tony Lin last updated on 01/Feb/20
let IE=x  ((x^2 +x^2 −1^2 )/(2x^2 ))=cos((5π)/7)  let 2x^2 =t  ⇒((t−1)/t)=cos((5π)/7)  t=(1/(1−cos((5π)/7)))  IH=(√(x^2 +x^2 −2x^2 cos((3π)/7)))        =(√(t(1−cos((3π)/7))))        =(√((1−cos((3π)/7))/(1−cos((5π)/7))))
letIE=xx2+x2122x2=cos5π7let2x2=tt1t=cos5π7t=11cos5π7IH=x2+x22x2cos3π7=t(1cos3π7)=1cos3π71cos5π7
Commented by M±th+et£s last updated on 01/Feb/20
nice work sir
niceworksir
Answered by mr W last updated on 01/Feb/20
Commented by mr W last updated on 01/Feb/20
α=(π/7)  AP=AB×sin α=sin α  PL=((AP)/(tan 2α))=((sin α)/(tan 2α))  HI=KL=2PL=((2sin α)/(tan 2α))=((2 sin (π/7))/(tan ((2π)/7)))≈0.692
α=π7AP=AB×sinα=sinαPL=APtan2α=sinαtan2αHI=KL=2PL=2sinαtan2α=2sinπ7tan2π70.692
Commented by M±th+et£s last updated on 01/Feb/20
thank you sir
thankyousir

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