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} \\ $$$$\:\:\mathrm{4cos}\:^{\mathrm{3}} \alpha−\mathrm{3cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{2cos}\:\alpha} \\ $$$$\:\:\mathrm{cos}\:\alpha=\:\frac{\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}}\: \\ $$$$\:\mathrm{sin}\:\alpha\:=\:\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{17}}}}{\mathrm{4}} \\…
Question Number 49384 by Pk1167156@gmail.com last updated on 06/Dec/18 Commented by Kunal12588 last updated on 06/Dec/18 $${for}\:\bigtriangleup{ABD} \\ $$$${AD}=\mathrm{39} \\ $$$${pythagorean}\:{triples}\:{with}\:{side}\:\mathrm{39} \\ $$$$\mathrm{1}.\:\mathrm{39},\mathrm{52},\mathrm{65}\:\:\:\:\:\:\:\:\:\:\:× \\ $$$$\mathrm{2}.\mathrm{39},\mathrm{80},\mathrm{89}\:\:\:…
Question Number 180458 by mnjuly1970 last updated on 12/Nov/22 Answered by mr W last updated on 13/Nov/22 $${c}^{\mathrm{2}} =\mathrm{1}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\mathrm{2}×\mathrm{1}×\sqrt{\mathrm{3}}×\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{4}+\sqrt{\mathrm{3}} \\ $$$$\frac{\mathrm{sin}\:\alpha}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{sin}\:\mathrm{120}°}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{sin}\:\alpha=\sqrt{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}=\frac{\mathrm{3}}{\mathrm{2}\sqrt{\mathrm{4}+\sqrt{\mathrm{3}}}}…
Question Number 114906 by mr W last updated on 21/Sep/20 Commented by MJS_new last updated on 22/Sep/20 $$\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{a}\:\mathrm{solution}?\:\mathrm{and}\:\mathrm{is}\:\mathrm{it}\:“\mathrm{nice}''? \\ $$$$\mathrm{I}\:\mathrm{get} \\ $$$${a}\approx\mathrm{6}.\mathrm{78750} \\ $$$${b}\approx\mathrm{7}.\mathrm{55680} \\…
Question Number 49365 by behi83417@gmail.com last updated on 06/Dec/18 Commented by behi83417@gmail.com last updated on 06/Dec/18 $${CA}=\mathrm{6},{AD}=\sqrt{\mathrm{3}},{DE}=\sqrt{\mathrm{2}},{EF}=\mathrm{1} \\ $$$$\Rightarrow\:\:\:\:\:\:\boldsymbol{\mathrm{FC}}=? \\ $$ Answered by MJS last…
Question Number 49362 by behi83417@gmail.com last updated on 06/Dec/18 Commented by behi83417@gmail.com last updated on 06/Dec/18 $$\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{showen}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{picture}},\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{statement}} \\ $$$$\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{true}}? \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{FE}}^{\mathrm{2}} +\boldsymbol{\mathrm{FD}}^{\mathrm{2}} =\boldsymbol{\mathrm{BC}}^{\mathrm{2}} \\ $$…
Question Number 49360 by behi83417@gmail.com last updated on 06/Dec/18 Commented by behi83417@gmail.com last updated on 06/Dec/18 $$\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{of}}:{AGEC},\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{maximum}}, \\ $$$$\boldsymbol{\mathrm{then}}:\:\:\frac{\boldsymbol{\mathrm{GE}}}{\boldsymbol{\mathrm{AC}}}=? \\ $$ Commented by MJS last…
Question Number 180423 by mr W last updated on 12/Nov/22 Commented by Rasheed.Sindhi last updated on 12/Nov/22 $$\mathcal{I}\:{think}\:{sir}\:{that}\:{the}\:{area}\:{depends} \\ $$$${upon}\:{m}\overline {\mathrm{O}_{\mathrm{1}} \mathrm{O}_{\mathrm{2}} } \\ $$…