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Category: Geometry

Question-70519

Question Number 70519 by TawaTawa last updated on 05/Oct/19 Commented by mr W last updated on 05/Oct/19 $${B}+{D}=\mathrm{180}° \\ $$$$\frac{{B}}{{D}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${B}+\frac{\mathrm{3}}{\mathrm{2}}{B}=\mathrm{180} \\ $$$$\frac{\mathrm{5}}{\mathrm{2}}{B}=\mathrm{180} \\…

Question-70452

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-70429

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-135952

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}…

6-6-6-6-6-SOLUTION-let-x-6-6-6-6-6-therefore-x-2-6-6-6-6-6-the-equation-is-a-continuos-funtion-Thus-x-2-6-6-

Question Number 4847 by sanusihammed last updated on 17/Mar/16 $$\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$$$ \\ $$$${SOLUTION} \\ $$$$ \\ $$$${let}\:{x}\:=\:\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}+\sqrt{\mathrm{6}}}}}} \\ $$$$ \\ $$$${therefore}.. \\ $$$$ \\…

cos-cos-cos-cos-1-cos-1-sin-a-b-sin-a-sin-b-sin-1-a-sin-1-b-a-b-tan-a-a-b-k-tan-a-b-k-tan-1-a-b-k-

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-70277

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)…