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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
$${if}\:{ABCDEFG}\:{is}\:{a}\:{heptagon}\:{and}\:{AB}=\mathrm{1} \\ $$$${HIJKLM}\:{is}\:{hexagon}\: \\ $$$${than}\:{find}\:{HI} \\ $$
Commented by mr W last updated on 01/Feb/20
HIJKLMN is also a heptagon!
$${HIJKLMN}\:{is}\:{also}\:{a}\:{heptagon}! \\ $$
Commented by M±th+et£s last updated on 01/Feb/20
you are right sir its typo
$${you}\:{are}\:{right}\:{sir}\:{its}\:{typo} \\ $$
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))))
$${let}\:{IE}={x} \\ $$$$\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }={cos}\frac{\mathrm{5}\pi}{\mathrm{7}} \\ $$$${let}\:\mathrm{2}{x}^{\mathrm{2}} ={t} \\ $$$$\Rightarrow\frac{{t}−\mathrm{1}}{{t}}={cos}\frac{\mathrm{5}\pi}{\mathrm{7}} \\ $$$${t}=\frac{\mathrm{1}}{\mathrm{1}−{cos}\frac{\mathrm{5}\pi}{\mathrm{7}}} \\ $$$${IH}=\sqrt{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} {cos}\frac{\mathrm{3}\pi}{\mathrm{7}}} \\ $$$$\:\:\:\:\:\:=\sqrt{{t}\left(\mathrm{1}−{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}\right)} \\ $$$$\:\:\:\:\:\:=\sqrt{\frac{\mathrm{1}−{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}}{\mathrm{1}−{cos}\frac{\mathrm{5}\pi}{\mathrm{7}}}} \\ $$
Commented by M±th+et£s last updated on 01/Feb/20
nice work sir
$${nice}\:{work}\:{sir} \\ $$
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
$$\alpha=\frac{\pi}{\mathrm{7}} \\ $$$${AP}={AB}×\mathrm{sin}\:\alpha=\mathrm{sin}\:\alpha \\ $$$${PL}=\frac{{AP}}{\mathrm{tan}\:\mathrm{2}\alpha}=\frac{\mathrm{sin}\:\alpha}{\mathrm{tan}\:\mathrm{2}\alpha} \\ $$$${HI}={KL}=\mathrm{2}{PL}=\frac{\mathrm{2sin}\:\alpha}{\mathrm{tan}\:\mathrm{2}\alpha}=\frac{\mathrm{2}\:\mathrm{sin}\:\frac{\pi}{\mathrm{7}}}{\mathrm{tan}\:\frac{\mathrm{2}\pi}{\mathrm{7}}}\approx\mathrm{0}.\mathrm{692} \\ $$
Commented by M±th+et£s last updated on 01/Feb/20
thank you sir
$${thank}\:{you}\:{sir}\: \\ $$

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