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Question-189925




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

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