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