Question Number 48653 by ajfour last updated on 26/Nov/18 Commented by ajfour last updated on 26/Nov/18 $${Find}\:\:\frac{{r}}{{R}}.\:{Smaller}\:{sphere} \\ $$$$\:{inscribed}\:{inside}\:{tetrahedron} \\ $$$${and}\:{larger}\:{one}\:{circumscribing}\:{the} \\ $$$${tetrahedron}\:\left({regular}\right). \\ $$…
Question Number 48638 by mr W last updated on 26/Nov/18 Commented by mr W last updated on 26/Nov/18 $${Find}\:{the}\:{locus}\:{of}\:{P}\:{such}\:{that}\:{PO} \\ $$$${is}\:{the}\:{bisector}\:{of}\:\angle{APB}. \\ $$ Commented by…
Question Number 179688 by Acem last updated on 01/Nov/22 Commented by HeferH last updated on 01/Nov/22 Commented by HeferH last updated on 01/Nov/22 $$\phi=\mathrm{30}° \\…
Question Number 48611 by Tawa1 last updated on 25/Nov/18 Answered by mr W last updated on 26/Nov/18 $${r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ={radii}\:{of}\:{the}\:{two}\:{bigger}\:{circles} \\ $$$${r}_{\mathrm{3}} ={radius}\:{of}\:{the}\:{smallest}\:{circle} \\ $$$$\sqrt{\left({r}_{\mathrm{1}}…
Question Number 179653 by cherokeesay last updated on 31/Oct/22 Answered by mr W last updated on 31/Oct/22 Commented by mr W last updated on 31/Oct/22…
Question Number 48567 by behi83417@gmail.com last updated on 25/Nov/18 Commented by behi83417@gmail.com last updated on 25/Nov/18 $$\boldsymbol{\mathrm{square}}\:\boldsymbol{\mathrm{side}}=\mathrm{1},\boldsymbol{\mathrm{G}}:\boldsymbol{\mathrm{center}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{square}}. \\ $$$$\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{passes}}\:\boldsymbol{\mathrm{from}}:\boldsymbol{\mathrm{G}},\boldsymbol{\mathrm{F}},\boldsymbol{\mathrm{E}}. \\ $$$$\measuredangle\boldsymbol{\mathrm{GHE}}=\mathrm{120}^{\bullet} \\ $$$$\Rightarrow\boldsymbol{\mathrm{area}}\:\boldsymbol{\mathrm{of}}:\:\:\boldsymbol{\mathrm{GHEF}}=? \\ $$…
Question Number 48560 by behi83417@gmail.com last updated on 25/Nov/18 Commented by behi83417@gmail.com last updated on 25/Nov/18 $$\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{circle}}\:\boldsymbol{\mathrm{radi}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}=\:\boldsymbol{\mathrm{distance}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{midpoint}}\:\boldsymbol{\mathrm{of}}:\boldsymbol{\mathrm{AB}},\boldsymbol{\mathrm{BC}}\:\boldsymbol{\mathrm{from}} \\ $$$$\boldsymbol{\mathrm{out}}\:\boldsymbol{\mathrm{circle}}. \\ $$$$\boldsymbol{\mathrm{find}}:\:\boldsymbol{\mathrm{r}},\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{terms}}\:\boldsymbol{\mathrm{of}}:\:\boldsymbol{\mathrm{a}},\boldsymbol{\mathrm{b}}. \\ $$…
Question Number 179609 by Acem last updated on 31/Oct/22 Answered by mr W last updated on 31/Oct/22 $$\boldsymbol{{Method}}\:\boldsymbol{{I}} \\ $$$$\frac{\mathrm{sin}\:{x}}{\mathrm{2}\:\mathrm{sin}\:\mathrm{15}°}=\frac{\mathrm{sin}\:\left({x}+\mathrm{15}°\right)}{\mathrm{1}}=\frac{\mathrm{sin}\:{x}\:\mathrm{cos}\:\mathrm{15}°+\mathrm{cos}\:{x}\:\mathrm{sin}\:\mathrm{15}°}{\mathrm{1}} \\ $$$$\mathrm{sin}\:\mathrm{30}°+\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{15}°}{\mathrm{tan}\:{x}}=\mathrm{1} \\ $$$$\frac{\mathrm{1}−\mathrm{cos}\:\mathrm{30}°}{\mathrm{tan}\:\gamma}=\frac{\mathrm{1}}{\mathrm{2}}…
Question Number 48522 by ajfour last updated on 25/Nov/18 Commented by ajfour last updated on 25/Nov/18 $${A}\:{regular}\:{hexagonal}\:{pyramid} \\ $$$${with}\:{base}\:{edge}\:\boldsymbol{{a}}\:{and}\:{altitude}\:\boldsymbol{{h}}. \\ $$$${Find}\:{the}\:{area}\:{of}\:{a}\:{section}\:{that} \\ $$$${passes}\:{through}\:{midpoints}\:{P},\:{Q} \\ $$$${of}\:{sides}\:{AB}\:{and}\:{CD}\:{and}\:{also}…
Question Number 179567 by Acem last updated on 30/Oct/22 Answered by mr W last updated on 30/Oct/22 $${x}^{\mathrm{2}} =\left(\mathrm{3}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} =\mathrm{71} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{71}} \\…