Question Number 180623 by cortano1 last updated on 14/Nov/22 Answered by HeferH last updated on 14/Nov/22 $$\left({R}\:+\:{r}\right)^{\mathrm{2}} =\:{r}^{\mathrm{2}} \:+\:\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{2}{r}}{{R}}\:=\:\frac{\mathrm{1}\:+\:\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\: \\ $$…
Question Number 180615 by cherokeesay last updated on 14/Nov/22 Answered by Acem last updated on 14/Nov/22 Commented by cherokeesay last updated on 14/Nov/22 $${thank}\:{you}\:{sir}. \\…
Question Number 49536 by behi83417@gmail.com last updated on 07/Dec/18 Commented by behi83417@gmail.com last updated on 07/Dec/18 $$\bigtriangleup\:{and}\bigtriangleup\:{are}\:{equilateral}\:{triangles}\:{with} \\ $$$${parallel}\:{sides}\:{and}\:\boldsymbol{\mathrm{PM}},{is}\:{the}\:{same}\:{for} \\ $$$${all}\:{sides}.{if}:{area}\left(\bigtriangleup\right)=\mathrm{32}\:{and}\:{area}\left(\bigtriangleup\right)=\mathrm{6}, \\ $$$${then}:{valve}\:{of}\::\boldsymbol{\mathrm{PM}}=? \\ $$…
Question Number 49530 by behi83417@gmail.com last updated on 07/Dec/18 Commented by behi83417@gmail.com last updated on 07/Dec/18 $$\bigtriangleup\:{and}\:\bigtriangleup\:{are}\:{equilateral}\:{triangles}. \\ $$$${AB}=\mathrm{4},{EG}=\mathrm{2}\sqrt{\mathrm{2}}\:\:\:\left[{or}:{AB}=\mathrm{6},{EG}=\mathrm{2}\sqrt{\mathrm{6}}\right] \\ $$$$\Rightarrow\:\:\:\:\:{AE}=?\:\:\:\:,\:\:\:\:\:\measuredangle{AEI}=? \\ $$ Commented by…
Question Number 180582 by HeferH last updated on 14/Nov/22 Answered by mr W last updated on 14/Nov/22 Commented by mr W last updated on 14/Nov/22…
Question Number 180547 by mnjuly1970 last updated on 13/Nov/22 Answered by mr W last updated on 13/Nov/22 Commented by mr W last updated on 13/Nov/22…
Question Number 180510 by cherokeesay last updated on 13/Nov/22 Commented by JDamian last updated on 13/Nov/22 Great roundabout expression for "yellow rectangle" Answered by MJS_new last updated on 13/Nov/22 $$\left(\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}}…
Question Number 49433 by behi83417@gmail.com last updated on 06/Dec/18 Commented by ajfour last updated on 06/Dec/18 $${circle}\:{radius},\:{Sir}\:? \\ $$ Commented by behi83417@gmail.com last updated on…
Question Number 49430 by behi83417@gmail.com last updated on 06/Dec/18 Commented by behi83417@gmail.com last updated on 07/Dec/18 $${D}\overset{} {{A}E}=\boldsymbol{\theta},{AD}=\boldsymbol{{r}}\:\:\:. \\ $$$${there}\:{is}\:{two}\:{equilateral}\:{triangles}, \\ $$$$\:{inscribed}\:{in}\:{circle}\:,{one}\:{vertex}\:{on}\: \\ $$$${point}:{C}\:{and}\:{two}\:{anothers}\:{locate}\:{on} \\…
Question Number 180459 by mnjuly1970 last updated on 12/Nov/22 Answered by greougoury555 last updated on 12/Nov/22 $$\:\:\mathrm{cos}\:\mathrm{3}\alpha=\frac{\mathrm{1}}{\mathrm{2cos}\:\alpha} \\ $$$$\:\:\mathrm{4cos}\:^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{2cos}\:\alpha} \\ $$$$\:\:\mathrm{cos}\:\alpha=\:\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}}\: \\ $$$$\:\mathrm{sin}\:\alpha\:=\:\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}} \\…