Question Number 210172 by efronzo1 last updated on 01/Aug/24 Answered by a.lgnaoui last updated on 02/Aug/24 $$\frac{\mathrm{sin}\:\mathrm{2}\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{BE}}}=\frac{\mathrm{cos}\:\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{AB}}}\:\:\:\:\left(\measuredangle\mathrm{AED}=\mathrm{90}−\mathrm{2a}\right) \\ $$$$\mathrm{AB}=\frac{\mathrm{DEcos}\:\boldsymbol{\mathrm{a}}}{\mathrm{sin}\:\mathrm{2}\boldsymbol{\mathrm{a}}}.=\frac{\boldsymbol{\mathrm{BE}}}{\mathrm{2sin}\:\boldsymbol{\mathrm{a}}} \\ $$$$\boldsymbol{\mathrm{Aire}}\left(\boldsymbol{\mathrm{ABC}}\right)=\boldsymbol{\mathrm{Air}}\mathrm{e}\left(\boldsymbol{\mathrm{ADFC}}\right)+\boldsymbol{\mathrm{Aire}}\left(\boldsymbol{\mathrm{DBF}}\right) \\ $$$$\boldsymbol{\mathrm{A}}\mathrm{ire}\left(\mathrm{BDE}\right)=\mathrm{Aire}\left(\boldsymbol{\mathrm{BEF}}\right)+\boldsymbol{\mathrm{Aire}}\left(\boldsymbol{\mathrm{BDF}}\right) \\ $$$$\Rightarrow\:…
Question Number 210120 by emilagazade last updated on 31/Jul/24 Answered by mr W last updated on 31/Jul/24 Commented by mr W last updated on 31/Jul/24…
Question Number 210122 by SonGoku last updated on 31/Jul/24 Commented by SonGoku last updated on 31/Jul/24 $$\mathrm{How}\:\mathrm{do}\:\mathrm{I}\:\mathrm{find}\:\mathrm{the}\:\mathrm{painted}\:\mathrm{area}? \\ $$ Answered by mr W last updated…
Question Number 210090 by Tawa11 last updated on 30/Jul/24 Commented by mr W last updated on 30/Jul/24 $${are}\:{there}\:\mathrm{3}\:{times}\:{letter}\:{p}\:{in}\:{your}\: \\ $$$${picture}? \\ $$ Commented by Tawa11…
Question Number 209965 by mnjuly1970 last updated on 27/Jul/24 Answered by mr W last updated on 27/Jul/24 Commented by mr W last updated on 27/Jul/24…
Question Number 209932 by cherokeesay last updated on 26/Jul/24 Answered by mahdipoor last updated on 26/Jul/24 $${BD}^{\mathrm{2}} =\mathrm{100}+\mathrm{49}−\mathrm{140}{cosC}=\mathrm{121}+\mathrm{64}−\mathrm{176}{cosA} \\ $$$$\frac{{BD}}{{sinC}}=\frac{{BD}}{{sinA}}=\mathrm{2}{R}\:\Rightarrow{C}+{A}=\mathrm{180} \\ $$$$\Rightarrow\mathrm{149}−\mathrm{140}{cosC}=\mathrm{185}−\mathrm{176}{cosA} \\ $$$$\Rightarrow{cosC}=\frac{\mathrm{185}−\mathrm{149}}{−\mathrm{176}−\mathrm{140}}=−\frac{\mathrm{9}}{\mathrm{79}} \\…
Question Number 209937 by Ari last updated on 26/Jul/24 Answered by mr W last updated on 26/Jul/24 Commented by mr W last updated on 26/Jul/24…
Question Number 209913 by mr W last updated on 25/Jul/24 Commented by mr W last updated on 25/Jul/24 $${AB}\:{is}\:{diameter},\:{CD}\:{is}\:{tangent}.\: \\ $$$${AE}={ED},\:{AB}//{CD} \\ $$$${find}\:{angle}\:{x}=? \\ $$…
Question Number 209889 by Tawa11 last updated on 24/Jul/24 Answered by som(math1967) last updated on 25/Jul/24 $${let}\:{PQ}={QS}={QR}={x} \\ $$$${cos}\angle{PQS}=\frac{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{12}^{\mathrm{2}} }{\mathrm{2}{x}^{\mathrm{2}} }=\frac{{x}^{\mathrm{2}} −\mathrm{72}}{{x}^{\mathrm{2}} }…
Question Number 209813 by mr W last updated on 22/Jul/24 Answered by mahdipoor last updated on 22/Jul/24 $${R}=\:{radius}\:{red}\:{circle} \\ $$$${r}\:=\:{radius}\:{white}\:{circle} \\ $$$$\begin{cases}{\mathrm{2}{R}=\mathrm{8}+\mathrm{2}+\mathrm{2}{r}}\\{{C}_{\mathrm{1}} {C}_{\mathrm{2}} =\sqrt{{r}^{\mathrm{2}} +\left({R}−\left({r}+\mathrm{2}\right)\right)^{\mathrm{2}}…