Question-194029

Question Number 194029 by Rupesh123 last updated on 26/Jun/23

Commented by Rupesh123 last updated on 26/Jun/23

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Answered by Subhi last updated on 26/Jun/23

AD = (√(2Q^2 −2Q^2 cos(120)))=(√3)Q AB = AC = Z (Z/(sin(150)))=(((√3)Q)/(sin(20))) ⇛ Q = ((2sin(20)Z)/( (√3))) ((2sin(20)Z)/( (√3).sin(120−x)))=((z−((2sin(120)Z)/( (√3))))/(sin(x))) 2sin(20)sin(x)=((((√3)cos(x))/2)+((sin(x))/2))((√3)−2sin(20)) x = tan^(−1) (((((√3)/2)((√3)−2sin(20)))/(2sin(20)−(((√3)−2sin(20))/2))))=80

$${AD}\:=\:\sqrt{\mathrm{2}{Q}^{\mathrm{2}} −\mathrm{2}{Q}^{\mathrm{2}} {cos}\left(\mathrm{120}\right)}=\sqrt{\mathrm{3}}{Q} \\ $$$${AB}\:=\:{AC}\:=\:{Z} \\ $$$$\frac{{Z}}{{sin}\left(\mathrm{150}\right)}=\frac{\sqrt{\mathrm{3}}{Q}}{{sin}\left(\mathrm{20}\right)}\:\Rrightarrow\:{Q}\:=\:\frac{\mathrm{2}{sin}\left(\mathrm{20}\right){Z}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{\mathrm{2}{sin}\left(\mathrm{20}\right){Z}}{\:\sqrt{\mathrm{3}}.{sin}\left(\mathrm{120}−{x}\right)}=\frac{{z}−\frac{\mathrm{2}{sin}\left(\mathrm{120}\right){Z}}{\:\sqrt{\mathrm{3}}}}{{sin}\left({x}\right)} \\ $$$$\mathrm{2}{sin}\left(\mathrm{20}\right){sin}\left({x}\right)=\left(\frac{\sqrt{\mathrm{3}}{cos}\left({x}\right)}{\mathrm{2}}+\frac{{sin}\left({x}\right)}{\mathrm{2}}\right)\left(\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{20}\right)\right) \\ $$$${x}\:=\:{tan}^{−\mathrm{1}} \left(\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{20}\right)\right)}{\mathrm{2}{sin}\left(\mathrm{20}\right)−\frac{\sqrt{\mathrm{3}}−\mathrm{2}{sin}\left(\mathrm{20}\right)}{\mathrm{2}}}\right)=\mathrm{80} \\ $$

Commented by Subhi last updated on 26/Jun/23

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