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Question-193906




Question Number 193906 by cherokeesay last updated on 22/Jun/23
Answered by Subhi last updated on 22/Jun/23
  (1/(sin(18)))=((1+y)/(sin(144−x)))   (1/(sin(18)))=(y/(sin(x))) ⇛ y = ((sin(x))/(sin(18)))  (1/(sin(18)))=((1+((sin(x))/(sin(18))))/(sin(144−x)))  sin(144−x)=sin(18)+sin(x)  sin(144)cos(x)−cos(144)sin(x)=sin(18)+sin(x)  sin(144)cos(x)+(−cos(144)−1)sin(x)=sin(18)  put sin(x) = u  sin(144).(√(1−u^2 )) = (1+cos(144))^2 u+sin(18)  (−sin^2 (144)−(1+cos(144))^2 )u^2 −2sin(18)(1+cos(144))u+(sin^2 (144)−sin^2 (18))=0  u = 0.66913  x = sin^(−1) (0.66913) = 42  (z/(sin(162−x)))=(1/(sin(18)))  z = ((sin(162−x))/(sin(18)))=2.8
$$ \\ $$$$\frac{\mathrm{1}}{{sin}\left(\mathrm{18}\right)}=\frac{\mathrm{1}+{y}}{{sin}\left(\mathrm{144}−{x}\right)}\: \\ $$$$\frac{\mathrm{1}}{{sin}\left(\mathrm{18}\right)}=\frac{{y}}{{sin}\left({x}\right)}\:\Rrightarrow\:{y}\:=\:\frac{{sin}\left({x}\right)}{{sin}\left(\mathrm{18}\right)} \\ $$$$\frac{\mathrm{1}}{{sin}\left(\mathrm{18}\right)}=\frac{\mathrm{1}+\frac{{sin}\left({x}\right)}{{sin}\left(\mathrm{18}\right)}}{{sin}\left(\mathrm{144}−{x}\right)} \\ $$$${sin}\left(\mathrm{144}−{x}\right)={sin}\left(\mathrm{18}\right)+{sin}\left({x}\right) \\ $$$${sin}\left(\mathrm{144}\right){cos}\left({x}\right)−{cos}\left(\mathrm{144}\right){sin}\left({x}\right)={sin}\left(\mathrm{18}\right)+{sin}\left({x}\right) \\ $$$${sin}\left(\mathrm{144}\right){cos}\left({x}\right)+\left(−{cos}\left(\mathrm{144}\right)−\mathrm{1}\right){sin}\left({x}\right)={sin}\left(\mathrm{18}\right) \\ $$$${put}\:{sin}\left({x}\right)\:=\:{u} \\ $$$${sin}\left(\mathrm{144}\right).\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }\:=\:\left(\mathrm{1}+{cos}\left(\mathrm{144}\right)\right)^{\mathrm{2}} {u}+{sin}\left(\mathrm{18}\right) \\ $$$$\left(−{sin}^{\mathrm{2}} \left(\mathrm{144}\right)−\left(\mathrm{1}+{cos}\left(\mathrm{144}\right)\right)^{\mathrm{2}} \right){u}^{\mathrm{2}} −\mathrm{2}{sin}\left(\mathrm{18}\right)\left(\mathrm{1}+{cos}\left(\mathrm{144}\right)\right){u}+\left({sin}^{\mathrm{2}} \left(\mathrm{144}\right)−{sin}^{\mathrm{2}} \left(\mathrm{18}\right)\right)=\mathrm{0} \\ $$$${u}\:=\:\mathrm{0}.\mathrm{66913} \\ $$$${x}\:=\:{sin}^{−\mathrm{1}} \left(\mathrm{0}.\mathrm{66913}\right)\:=\:\mathrm{42} \\ $$$$\frac{{z}}{{sin}\left(\mathrm{162}−{x}\right)}=\frac{\mathrm{1}}{{sin}\left(\mathrm{18}\right)} \\ $$$${z}\:=\:\frac{{sin}\left(\mathrm{162}−{x}\right)}{{sin}\left(\mathrm{18}\right)}=\mathrm{2}.\mathrm{8} \\ $$
Commented by Subhi last updated on 22/Jun/23

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