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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

$$ifABCDEFGisaheptagonandAB=1$$$$HIJKLMishexagon$$$$thanfindHI$$

Commented by mr W last updated on 01/Feb/20

$$HIJKLMNisalsoaheptagon!$$

Commented by M±th+et£s last updated on 01/Feb/20

$$youarerightsiritstypo$$

Commented by Tony Lin last updated on 01/Feb/20

$$letIE=x$$$$\frac{x^2+x^2-1^2}{2x^2}=cos\frac{5\pi }{7}$$$$let2x^2=t$$$$\Rightarrow \frac{t-1}{t}=cos\frac{5\pi }{7}$$$$t=\frac{1}{1-cos\frac{5\pi }{7}}$$$$IH=\sqrt{x^2+x^2-2x^{2}cos\frac{3\pi }{7}}$$$$=\sqrt{t(1-cos\frac{3\pi }{7})}$$$$=\sqrt{\frac{1-cos\frac{3\pi }{7}}{1-cos\frac{5\pi }{7}}}$$

Commented by M±th+et£s last updated on 01/Feb/20

$$niceworksir$$

Answered by mr W last updated on 01/Feb/20

Commented by mr W last updated on 01/Feb/20

$$\alpha =\frac{\pi }{7}$$$$AP=AB\times \sin \alpha =\sin \alpha$$$$PL=\frac{AP}{\tan 2\alpha }=\frac{\sin \alpha }{\tan 2\alpha }$$$$HI=KL=2PL=\frac{2\sin \alpha }{\tan 2\alpha }=\frac{2\sin \frac{\pi }{7}}{\tan \frac{2\pi }{7}}\approx 0.692$$

Commented by M±th+et£s last updated on 01/Feb/20

$$thankyousir$$

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