Question Number 70497 by mr W last updated on 04/Oct/19 Commented by mr W last updated on 04/Oct/19 $${shaded}\:{area}=? \\ $$ Answered by ajfour last…
Question Number 70452 by ajfour last updated on 04/Oct/19 Commented by ajfour last updated on 04/Oct/19 $${ABCD}\:{is}\:{a}\:{square}.\:{Find}\:{minimum} \\ $$$${side}\:{length}\:\boldsymbol{{s}}\:{of}\:{equilateral}\:\bigtriangleup{DEF} \\ $$$${in}\:{terms}\:{of}\:{quarter}\:{circle}\:{radii} \\ $$$${a}\:{and}\:{b}.\:{Also}\:{find}\:{range}\:{of}\: \\ $$$$\:\frac{{a}}{{b}}\:{so}\:{that}\:{such}\:{an}\:{equilateral}…
Question Number 70429 by TawaTawa last updated on 04/Oct/19 Commented by TawaTawa last updated on 04/Oct/19 If ABCD is a square, AH is the tangent of the sector BEF at G, AG:GH = 7, and DH = 2. Find the area of the square Commented by MJS last updated on 04/Oct/19 $${AG}:{GH}=\mathrm{7}???…
Question Number 135952 by Dwaipayan Shikari last updated on 17/Mar/21 Commented by Dwaipayan Shikari last updated on 17/Mar/21 $${It}\:{isArchimedes}'{s}\:{Nothing}\:{Grinder}.\:{We}\:{can}\:{move}\:{this}\:{back} \\ $$$${and}\:{forth}\:{and}\:{sideways}\:.{Prove}\:{that}\:{the}\:{tip}\:{on}\:{various} \\ $$$${positions}\:{on}\:{that}\:{hand}\:{of}\:{the}\:{wodden}\:{grinder}\:{creates}\: \\ $$$${various}\:{shapes}\:{of}\:{ellipses}…
Question Number 4847 by sanusihammed last updated on 17/Mar/16 $$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$$$ \\ $$$${SOLUTION} \\ $$$$ \\ $$$${let}\:{x}\:=\:\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$$$ \\ $$$${therefore}.. \\ $$$$ \\…
Question Number 4783 by Dnilka228 last updated on 10/Mar/16 $$\mathrm{cos}\:\alpha+\beta\:\approx\left(\frac{\mathrm{cos}\:\alpha+\mathrm{cos}\:\beta}{\mathrm{cos}^{−\mathrm{1}} \alpha+\mathrm{cos}^{−\mathrm{1}} \beta}\right)^{\alpha+\beta} \\ $$$$\mathrm{sin}\:{a}+{b}\approx\left(\frac{\mathrm{sin}\:{a}+\mathrm{sin}\:{b}}{\mathrm{sin}^{−\mathrm{1}} {a}+\mathrm{sin}^{−\mathrm{1}} {b}}\right)^{{a}+{b}} \\ $$$$\mathrm{tan}\:\left({a}+\frac{{a}}{{b}}\right)^{{k}} \approx\left(\frac{\mathrm{tan}\:\left({a}+{b}\right)×{k}}{\mathrm{tan}^{−\mathrm{1}} \left({a}+{b}\right)×{k}}\right) \\ $$ Commented by Dnilka228…
Question Number 70277 by mr W last updated on 02/Oct/19 Answered by mind is power last updated on 02/Oct/19 $$\frac{{a}}{{sin}\left(\theta\right)}=\frac{\mathrm{2}{a}}{{Sin}\left(\pi−\mathrm{3}\theta\right)}=\frac{\mathrm{2}{a}}{{sin}\left(\mathrm{3}\theta\right)} \\ $$$$\Rightarrow{sin}\left(\mathrm{3}\theta\right)=\mathrm{2}{sin}\left(\theta\right) \\ $$$${sin}\left(\mathrm{3}\theta\right)=−\mathrm{4}{sin}^{\mathrm{3}} \left(\theta\right)+\mathrm{3}{sin}\left(\theta\right)…
Question Number 4730 by Rojaye Shegz last updated on 01/Mar/16 $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dx}}{{sinx}^{{cosx}\:} \:\:+\:\:\:\:{cosx}^{{sinx}\:} } \\ $$ Commented by prakash jain last updated on 01/Mar/16…
Question Number 70256 by behi83417@gmail.com last updated on 02/Oct/19 Commented by behi83417@gmail.com last updated on 02/Oct/19 $$\mathrm{ABCD},\mathrm{is}\:\mathrm{a}\:\mathrm{squre}. \\ $$$$\mathrm{F},\mathrm{is}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{AC}. \\ $$$$\mathrm{1}.\mathrm{if}:\:\:\:\boldsymbol{\mathrm{DF}}=\mathrm{4},\boldsymbol{\mathrm{EB}}=\mathrm{3},\mathrm{area}\:\mathrm{of}:\boldsymbol{\mathrm{ACE}}=? \\ $$$$\mathrm{2}.\mathrm{if}:\boldsymbol{\mathrm{AC}},\mathrm{be}\:\mathrm{diagonal},\mathrm{but}\:\mathrm{AF}\neq\mathrm{FC}\:\mathrm{and}: \\ $$$$\mathrm{DF}=\mathrm{4},\mathrm{FE}=\mathrm{2},\mathrm{EB}=\mathrm{3},\mathrm{now}\:\mathrm{find}\:\mathrm{area}\:\mathrm{of}:\mathrm{ACE}.…
Question Number 70225 by TawaTawa last updated on 02/Oct/19 Commented by TawaTawa last updated on 02/Oct/19 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\:\mathrm{A}\left(\mathrm{ECD}\right)\:−\:\mathrm{A}\left(\mathrm{BAE}\right)\:\:\mathrm{in}\:\mathrm{cm}^{\mathrm{2}} \\ $$ Commented by MJS last updated on…