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 200636 by a.lgnaoui last updated on 21/Nov/23

$$1-\mathrm{Determiner}\mathrm{la}\mathrm{valeur}\mathrm{de}\mathbf{EF}$$$$2-\mathrm{Laire}\mathrm{du}\mathrm{triangle}\mathbf{ADE}$$$$$$

Commented by a.lgnaoui last updated on 21/Nov/23

Commented by witcher3 last updated on 21/Nov/23

$$\left(\mathrm{ED}\right)\mathrm{\perp }\left(\mathrm{BD}\right)..?$$$$\mathrm{and}\left(\mathrm{AB}\right)\mathrm{\perp }\left(\mathrm{AC}\right)\mathrm{or}\mathrm{AE}...$$$$$$

Commented by a.lgnaoui last updated on 21/Nov/23

$$\mathbf{BD}\mathrm{\perp }\mathrm{D}\mathbf{E}\mathbf{and}\mathrm{AB}\mathrm{\perp }\mathbf{A}\mathrm{C}$$

Commented by a.lgnaoui last updated on 21/Nov/23

$$\measuredangle \mathbf{C}=30$$

Commented by witcher3 last updated on 21/Nov/23

$$\mathrm{AC}=\frac{6}{\mathrm{tg}\left(30\right)}=6\sqrt{3}$$$$\mathrm{AD}=6\sqrt{3}-8$$$$\mathrm{BD}=\sqrt{6^2+{(6\sqrt{3}-8)}^2}$$$$\mathrm{\angle }\mathrm{EDB}=60-{\tan }^{-1}\left(\frac{6\sqrt{3}-8}{6}\right)=\theta$$$$\mathrm{BE}=\sqrt{{\mathrm{BD}}^2+{\mathrm{DE}}^2},\mathrm{DE}=\mathrm{BDtan}\left(\theta \right)$$$${\mathrm{EA}}^2={\mathrm{BE}}^2+{\mathrm{AB}}^2-2\mathrm{BE}.\mathrm{ABcos}\left(60\right)$$$$\mathrm{we}\mathrm{know}\mathrm{evrey}\mathrm{length}..\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{side}\mathrm{of}\mathrm{triangle}$$$$\mathrm{S}=\sqrt{\frac{\mathrm{p}}{2}(\frac{\mathrm{p}}{2}-\mathrm{a})(\frac{\mathrm{p}}{2}-\mathrm{b})(\frac{\mathrm{p}}{2}-\mathrm{c})}$$$$\mathrm{P}=\mathrm{a}+\mathrm{b}+\mathrm{c},$$

Answered by mr W last updated on 21/Nov/23

$$AC=6\sqrt{3}$$$$AD=6\sqrt{3}-8$$$$BD=\sqrt{6^2+{(6\sqrt{3}-8)}^2}=4\sqrt{13-6\sqrt{3}}$$$$\mathrm{\angle }ABD={\tan }^{-1}\frac{6\sqrt{3}-8}{6}$$$$DE=BD\tan (60°-\mathrm{\angle }ABD)$$$$=4\sqrt{13-6\sqrt{3}}\times \frac{\sqrt{3}-\frac{6\sqrt{3}-8}{6}}{1+\sqrt{3}\times \frac{6\sqrt{3}-8}{6}}$$$$=\frac{4\sqrt{13-6\sqrt{3}}}{3-\sqrt{3}}$$$$\mathrm{\angle }CDE=\mathrm{\angle }ABD={\tan }^{-1}\frac{6\sqrt{3}-8}{6}$$$${\mathrm{\Delta }}_{ADE}=\frac{AD\times DE\times \sin \mathrm{\angle }CDE}{2}$$$$=\frac{6\sqrt{3}-8}{2}\times \frac{4\sqrt{13-6\sqrt{3}}}{3-\sqrt{3}}\times \frac{6\sqrt{3}-8}{\sqrt{{(6\sqrt{3}-8)}^2+6^2}}$$$$=\frac{57-29\sqrt{3}}{3}\approx 2.256842$$$$BE=\frac{BD}{\cos (60°-\mathrm{\angle }ABD)}=\frac{42-10\sqrt{3}}{3}$$$$EC=12-\frac{42-10\sqrt{3}}{3}=\frac{10\sqrt{3}-6}{3}$$$$\tan \mathrm{\angle }CAE=\frac{\frac{10\sqrt{3}-6}{3}\times \frac{1}{2}}{6\sqrt{3}-\frac{10\sqrt{3}-6}{3}\times \frac{\sqrt{3}}{2}}=\frac{45+2\sqrt{3}}{183}$$$$x=FE=\frac{DE}{\cos (\mathrm{\angle }CDE-\mathrm{\angle }CAE)}$$$$=\frac{DE}{\cos (\mathrm{\angle }ABD-\mathrm{\angle }CAE)}$$$$=\frac{4\sqrt{13-6\sqrt{3}}}{3-\sqrt{3}}\times \frac{1}{\frac{6}{\sqrt{6^2+{(6\sqrt{3}-8)}^2}}\times \frac{183}{\sqrt{{183}^2+{(45+2\sqrt{3})}^2}}+\frac{6\sqrt{3}-8}{\sqrt{6^2+{(6\sqrt{3}-8)}^2}}\times \frac{45+2\sqrt{3}}{\sqrt{{183}^2+{(45+2\sqrt{3})}^2}}}$$$$=\frac{4\sqrt{13-6\sqrt{3}}}{3-\sqrt{3}}\times \frac{\sqrt{458598-210816\sqrt{3}}}{774+254\sqrt{3}}$$$$=\frac{244\sqrt{2622-1476\sqrt{3}}}{390-3\sqrt{3}}\approx 5.131542$$

Commented by mr W last updated on 21/Nov/23

Terms of Service

Privacy Policy

Contact: info@tinkutara.com