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