Question Number 189925 by ajfour last updated on 24/Mar/23
Commented by ajfour last updated on 24/Mar/23
$${Regular}\:{pentagon}.\:{Find}\:{a}/{b}. \\ $$
Answered by BaliramKumar last updated on 24/Mar/23
$$\frac{\mathrm{5}+\sqrt{\mathrm{5}}}{\mathrm{3}+\sqrt{\mathrm{5}}}\:=\:\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$
Answered by HeferH last updated on 24/Mar/23
$$\mathrm{sin}\:\mathrm{18}°\:=\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:=\:\frac{\mathrm{2b}−\mathrm{a}}{\mathrm{2b}} \\ $$$$\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\:=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{a}}{\mathrm{b}} \\ $$$$\:\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}=\mathrm{2}\:−\frac{\mathrm{a}}{\mathrm{b}} \\ $$$$\:\frac{\mathrm{a}}{\mathrm{b}}\:=\:\frac{\mathrm{5}−\sqrt{\mathrm{5}}}{\mathrm{2}}\: \\ $$
Commented by HeferH last updated on 24/Mar/23