Question Number 113343 by mohammad17 last updated on 12/Sep/20
$${find}\:{the}\:{angle}\:{between}\:{x}+\mathrm{3}\sqrt{\mathrm{3}{y}}=\mathrm{2},\sqrt{\mathrm{3}{x}}−\mathrm{5}{y}=\mathrm{2} \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$
Commented by mr W last updated on 12/Sep/20
$${you}\:{have}\:{asked}\:\left({almost}\:{the}\:{same}\right) \\ $$$${question}\:{several}\:{times},\:{e}.{g}.\:{Q}\mathrm{112902}, \\ $$$${they}\:{are}\:{all}\:{answered}\:{in}\:{detail}. \\ $$
Answered by Dwaipayan Shikari last updated on 12/Sep/20
$${x}+\mathrm{3}\sqrt{\mathrm{3}}{y}=\mathrm{2} \\ $$$${y}=−\frac{{x}}{\mathrm{3}\sqrt{\mathrm{3}}}+\mathrm{2} \\ $$$${y}=\frac{\sqrt{\mathrm{3}}}{\mathrm{5}}{x}−\mathrm{2} \\ $$$${tan}\theta=\mid\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{3}}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{15}}}\mid=\mid\frac{\frac{\mathrm{14}}{\mathrm{15}\sqrt{\mathrm{3}}}}{\frac{\mathrm{14}}{\mathrm{15}}}\mid=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\theta=\frac{\pi}{\mathrm{6}} \\ $$