Category: Geometry

Question Number 43834 by MrW3 last updated on 16/Sep/18 Commented by MrW3 last updated on 16/Sep/18 $$Anewsolutionforanoldquestion:$$ Commented by MrW3 last updated on…

Question Number 174888 by ajfour last updated on 13/Aug/22 Answered by aleks041103 last updated on 13/Aug/22 $$ifOiscenterofcircle:$$$$\mathrm{\angle }AOB=\mathrm{\angle }QOP\Rightarrow \stackrel{⌢}{AB}=\stackrel{⌢}{PQ}$$$$\Rightarrow AB=PQ$$$${By}\:{analogy}:…

Question Number 174763 by dragan91 last updated on 10/Aug/22 Answered by a.lgnaoui last updated on 11/Aug/22 $$\mathrm{\angle }ADF$$$$\frac{\sin \mathrm{A}}{\mathrm{DF}}=\frac{\sin \mathrm{F}}{\mathrm{AD}}=\frac{\sin \mathrm{D}}{24}\left(1\right)$$$$\mathrm{\angle }EBF$$$$\frac{\sin \mathrm{B}}{\mathrm{x}}=\frac{\sin \mathrm{F}}{3}=\frac{\sin \mathrm{E}}{8}\left(2\right)$$$$\angle{CDE}…

Question Number 43621 by ajfour last updated on 12/Sep/18 Commented by math1967 last updated on 13/Sep/18 $$AD=\sqrt{x^2-1},AC=\sqrt{x^2+x^2-1}=\sqrt{2x^2-1}$$$$\bigtriangleup{ABD}\sim\bigtriangleup{CED}\:\therefore\frac{{ED}}{{x}}=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \…

Question Number 174599 by vemulamanasa last updated on 05/Aug/22 Terms of Service Privacy Policy Contact: info@tinkutara.com