Question Number 182075 by Acem last updated on 04/Dec/22 Commented by Acem last updated on 04/Dec/22 $${Regular}\:{hexagon}\:\uparrow \\ $$ Answered by mr W last updated…
Question Number 182066 by HeferH last updated on 03/Dec/22 Answered by a.lgnaoui last updated on 04/Dec/22 $$\bigtriangleup\mathrm{ABC}\:\: \\ $$$$\mathrm{ABcos}\:\mathrm{2x}+\mathrm{BCcos}\:\mathrm{x}=\mathrm{ADcos}\:\mathrm{3x}+\mathrm{CDcos}\:\mathrm{4x} \\ $$$$\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{BC}}=\frac{\mathrm{sinx}\:}{\mathrm{AB}}\:\:\:\left(\mathrm{1}\right)\:\:\:\:\:\:\:\frac{\mathrm{sin}\:\mathrm{3x}}{\mathrm{CD}}\:\:\:\:=\frac{\mathrm{sin}\:\mathrm{4x}}{\mathrm{AD}}\left(\mathrm{2}\right) \\ $$$$\mathrm{ABsin}\:\mathrm{2x}=\mathrm{BCsin}\:\mathrm{x}\:\:\:\:\:\mathrm{AD}\:\mathrm{sin}\:\mathrm{3x}=\mathrm{CDsin}\:\mathrm{4x} \\ $$$$\mathrm{AB}=\frac{\mathrm{BCsin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{2x}}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{CD}=\frac{\mathrm{ADsin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}…
Question Number 182006 by mr W last updated on 03/Dec/22 Commented by mr W last updated on 03/Dec/22 $${find}\:{the}\:{largest}\:{area}\:{of}\:{an}\:{inscribed} \\ $$$${right}−{angle}\:{triangle}\:{in}\:{a}\:{given}\:{acute} \\ $$$${triangle}\:{with}\:{sides}\:{a},{b},{c}.\:\left({a}\geqslant{b}\geqslant{c}\right) \\ $$…
Question Number 116466 by mr W last updated on 04/Oct/20 Commented by mr W last updated on 04/Oct/20 $${R}={radius}\:{of}\:{big}\:{circles} \\ $$$${r}={radius}\:{of}\:{small}\:{circles} \\ $$$${let}\:\lambda=\frac{{r}}{{R}} \\ $$$$\mathrm{cos}\:\alpha=\frac{{R}}{{R}+{r}}=\frac{\mathrm{1}}{\mathrm{1}+\lambda}…
Question Number 182001 by mr W last updated on 03/Dec/22 Commented by mr W last updated on 03/Dec/22 $${find}\:{the}\:{area}\:{of}\:{triangle}. \\ $$ Answered by som(math1967) last…
Question Number 181987 by Acem last updated on 03/Dec/22 Answered by Frix last updated on 03/Dec/22 $$\left.\frac{\mathrm{sin}\:\mathrm{45}°}{{BC}}=\frac{\mathrm{sin}\:\mathrm{105}°}{{AC}}=\frac{\mathrm{sin}\:\mathrm{30}°}{{AB}}\right) \\ $$$$\Rightarrow\frac{{BC}}{{AC}}=\sqrt{\mathrm{3}}−\mathrm{1} \\ $$$${AC}+{BC}=\mathrm{14}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{get} \\ $$$${AC}=\frac{\mathrm{14}\sqrt{\mathrm{3}}}{\mathrm{3}}\wedge{BC}=\frac{\mathrm{14}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)}{\mathrm{3}}\:\Rightarrow\:{AB}=\frac{\mathrm{7}\sqrt{\mathrm{2}}\left(\mathrm{3}−\sqrt{\mathrm{3}}\right)}{\mathrm{3}} \\ $$…
Question Number 181976 by mr W last updated on 02/Dec/22 Answered by HeferH last updated on 02/Dec/22 $$\mathrm{5}? \\ $$ Commented by mr W last…
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Question Number 181935 by Acem last updated on 02/Dec/22 Answered by HeferH last updated on 02/Dec/22 Commented by Acem last updated on 04/Dec/22 $${Yes}\:{Sir}!\:\:{Thanksss}! \\…
Question Number 181918 by HeferH last updated on 02/Dec/22 Answered by Acem last updated on 02/Dec/22 Commented by Acem last updated on 02/Dec/22 $${Sir}\:{HeferH},\:{share}\:{with}\:{me}\:{to}\:{find}\:{x}\:{cause} \\…