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Question-193609

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Question Number 193609 by Mingma last updated on 17/Jun/23
Answered by Subhi last updated on 17/Jun/23
(x/(sin(60)))=(z/(sin(120−y))) (z/(sin(60)))=(1/(sin(60−y))) ⇛ z=((sin(60))/(sin(60−y))) x=((sin^2 (60))/(sin(60−y)sin(120−y))) ((x+z)/(sin(120)))=(z/(sin(60−y))) ((((sin^2 (60))/(sin(120−y)sin(60−y)))+((sin(60))/(sin(60−y))))/(sin(120)))=((sin(60))/(sin^2 (60−y))) ((sin^2 (60).sin(60−y))/(sin(120−y)))+sin(60)sin(60−y)=sin(60)sin(120) (3/4)sin(120−y)=((√3)/2)sin(60−y)sin(120−y)+(3/4)sin(60−y) ((3(√3))/8)cos^2 (y)−((√3)/8)sin^2 (y)−(3/4)sin(y)=0 cos^2 (y)=1−sin^2 (y) ((√3)/2)sin^2 (y)+(3/4)sin(y)−((3(√3))/8)=0 sin(y)=(((√(15))−(√3))/4) ⇛ y = 32.36 x=(((sin^2 (60)))/(sin(60−y)sin(120−y)))=1.618=((1+(√5))/2) (the golden ratio)
$$\frac{{x}}{{sin}\left(\mathrm{60}\right)}=\frac{{z}}{{sin}\left(\mathrm{120}−{y}\right)}\: \\ $$$$\frac{{z}}{{sin}\left(\mathrm{60}\right)}=\frac{\mathrm{1}}{{sin}\left(\mathrm{60}−{y}\right)}\:\Rrightarrow\:{z}=\frac{{sin}\left(\mathrm{60}\right)}{{sin}\left(\mathrm{60}−{y}\right)} \\ $$$${x}=\frac{{sin}^{\mathrm{2}} \left(\mathrm{60}\right)}{{sin}\left(\mathrm{60}−{y}\right){sin}\left(\mathrm{120}−{y}\right)} \\ $$$$\frac{{x}+{z}}{{sin}\left(\mathrm{120}\right)}=\frac{{z}}{{sin}\left(\mathrm{60}−{y}\right)} \\ $$$$\frac{\frac{{sin}^{\mathrm{2}} \left(\mathrm{60}\right)}{{sin}\left(\mathrm{120}−{y}\right){sin}\left(\mathrm{60}−{y}\right)}+\frac{{sin}\left(\mathrm{60}\right)}{{sin}\left(\mathrm{60}−{y}\right)}}{{sin}\left(\mathrm{120}\right)}=\frac{{sin}\left(\mathrm{60}\right)}{{sin}^{\mathrm{2}} \left(\mathrm{60}−{y}\right)} \\ $$$$\frac{{sin}^{\mathrm{2}} \left(\mathrm{60}\right).{sin}\left(\mathrm{60}−{y}\right)}{{sin}\left(\mathrm{120}−{y}\right)}+{sin}\left(\mathrm{60}\right){sin}\left(\mathrm{60}−{y}\right)={sin}\left(\mathrm{60}\right){sin}\left(\mathrm{120}\right) \\ $$$$\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{120}−{y}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}\left(\mathrm{60}−{y}\right){sin}\left(\mathrm{120}−{y}\right)+\frac{\mathrm{3}}{\mathrm{4}}{sin}\left(\mathrm{60}−{y}\right) \\ $$$$\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}{cos}^{\mathrm{2}} \left({y}\right)−\frac{\sqrt{\mathrm{3}}}{\mathrm{8}}{sin}^{\mathrm{2}} \left({y}\right)−\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({y}\right)=\mathrm{0} \\ $$$${cos}^{\mathrm{2}} \left({y}\right)=\mathrm{1}−{sin}^{\mathrm{2}} \left({y}\right) \\ $$$$\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sin}^{\mathrm{2}} \left({y}\right)+\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({y}\right)−\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}=\mathrm{0} \\ $$$${sin}\left({y}\right)=\frac{\sqrt{\mathrm{15}}−\sqrt{\mathrm{3}}}{\mathrm{4}}\:\Rrightarrow\:{y}\:=\:\mathrm{32}.\mathrm{36} \\ $$$${x}=\frac{\left({sin}^{\mathrm{2}} \left(\mathrm{60}\right)\right)}{{sin}\left(\mathrm{60}−{y}\right){sin}\left(\mathrm{120}−{y}\right)}=\mathrm{1}.\mathrm{618}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\left({the}\:{golden}\:{ratio}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by Subhi last updated on 17/Jun/23

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