Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 166063 by cortano1 last updated on 12/Feb/22

Commented by blackmamba last updated on 12/Feb/22

 tan α=(2/(6−2(√3)))=(1/(3−(√3)))=((3+(√3))/6)   tan 2α=((2(15+7(√3)))/(13))   tan (90°−2α)=(1/(tan 2α))=((15−7(√3))/(12))   m=((15−7(√3))/(12))

$$\:\mathrm{tan}\:\alpha=\frac{\mathrm{2}}{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\mathrm{3}−\sqrt{\mathrm{3}}}=\frac{\mathrm{3}+\sqrt{\mathrm{3}}}{\mathrm{6}} \\ $$$$\:\mathrm{tan}\:\mathrm{2}\alpha=\frac{\mathrm{2}\left(\mathrm{15}+\mathrm{7}\sqrt{\mathrm{3}}\right)}{\mathrm{13}} \\ $$$$\:\mathrm{tan}\:\left(\mathrm{90}°−\mathrm{2}\alpha\right)=\frac{\mathrm{1}}{\mathrm{tan}\:\mathrm{2}\alpha}=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$$$\:{m}=\frac{\mathrm{15}−\mathrm{7}\sqrt{\mathrm{3}}}{\mathrm{12}} \\ $$

Commented by MaxiMaths last updated on 12/Feb/22

cool...

$$\mathrm{cool}... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com