Category: Geometry

Question Number 43834 by MrW3 last updated on 16/Sep/18 Commented by MrW3 last updated on 16/Sep/18 $${A}\:{new}\:{solution}\:{for}\:{an}\:{old}\:{question}: \\ $$ 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 $${if}\:{O}\:{is}\:{center}\:{of}\:{circle}: \\ $$$$\angle{AOB}=\angle{QOP}\Rightarrow\overset{\frown} {{AB}}=\overset{\frown} {{PQ}} \\ $$$$\Rightarrow{AB}={PQ} \\ $$$${By}\:{analogy}:…

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

Question Number 43621 by ajfour last updated on 12/Sep/18 Commented by math1967 last updated on 13/Sep/18 $${AD}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:,{AC}=\sqrt{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{1}}=\sqrt{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\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