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} \\ $$$$\Rightarrow{x}=\sqrt{\mathrm{71}} \\…
Question Number 179566 by Acem last updated on 30/Oct/22 Answered by a.lgnaoui last updated on 30/Oct/22 $$\mathrm{AC}\mid\mid\mathrm{FD}\:\:\measuredangle\mathrm{BAC}=\mathrm{120}−\mathrm{90}=\mathrm{30} \\ $$$$\mathrm{sin}\:\mathrm{30}=\frac{\mathrm{BH}}{\mathrm{3}}\:\:\mathrm{BH}=\frac{\mathrm{3}}{\mathrm{2}}\:\:\Rightarrow\mathrm{BE}=\frac{\mathrm{3}}{\mathrm{2}}+\mathrm{5}=\frac{\mathrm{13}}{\mathrm{2}}\:\:\:\left(\mathrm{1}\right) \\ $$$$\measuredangle\mathrm{ABD}=\mathrm{360}−\mathrm{240}=\mathrm{120}\:\:\Rightarrow\measuredangle\mathrm{BHC}=\measuredangle\mathrm{BED}=\mathrm{90}−\mathrm{30}=\mathrm{60} \\ $$$$\:\mathrm{cos}\:\mathrm{60}=\frac{\mathrm{BE}}{\mathrm{BD}}\:\:\:\mathrm{BD}=\frac{\mathrm{BE}}{\mathrm{cos}\:\mathrm{60}}=\frac{\mathrm{13}}{\mathrm{2cos}\:\mathrm{60}} \\ $$$$\:\:\:\mathrm{BD}=\mathrm{13}…
Question Number 48466 by Tawa1 last updated on 24/Nov/18 Answered by tanmay.chaudhury50@gmail.com last updated on 24/Nov/18 $${A}\left(\mathrm{1},\mathrm{3}\right)\:\:{B}\left(\mathrm{7},\mathrm{9}\right)\:\:\:{D}\left(\mathrm{4},\mathrm{6}\right) \\ $$$${eqn}\:{AB}\:\:\left({y}−\mathrm{3}\right)=\frac{\mathrm{9}−\mathrm{3}}{\mathrm{7}−\mathrm{1}}\left({x}−\mathrm{1}\right) \\ $$$${y}−\mathrm{3}−{x}+\mathrm{1}=\mathrm{0} \\ $$$${y}−{x}−\mathrm{2}=\mathrm{0} \\ $$$${point}\:{C}\left(−\mathrm{2},\mathrm{0}\right)…