Question Number 191323 by Mingma last updated on 23/Apr/23 Answered by mr W last updated on 23/Apr/23 $$\frac{{A}}{{A}+{B}}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{{A}}{{A}+{B}+{C}}=\left(\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{9}} \\ $$$$\Rightarrow{A}=\frac{{A}+{B}+{C}}{\mathrm{9}} \\…
Question Number 191305 by Mingma last updated on 22/Apr/23 Commented by mr W last updated on 22/Apr/23 $${there}\:{is}\:{no}\:{unique}\:{solution}.\:{but}\:{there} \\ $$$${is}\:{a}\:{r}_{{min}} . \\ $$ Answered by…
Question Number 191278 by Mingma last updated on 22/Apr/23 Answered by a.lgnaoui last updated on 23/Apr/23 $$\mathrm{chaque}\:\mathrm{heptagone}\:\mathrm{a}\:\mathrm{7}\:\mathrm{triangles}\:\mathrm{isoceles}\:\mathrm{de}\:\mathrm{somet}\:\mathrm{O} \\ $$$$\mathrm{d}\:\mathrm{angle}\:\boldsymbol{\theta}=\frac{\mathrm{2}\pi}{\mathrm{7}}\:\:\mathrm{de}\:\mathrm{cote}\:\boldsymbol{\mathrm{x}}\:\:\mathrm{avec}\:\:\:\:\boldsymbol{\mathrm{x}}\mathrm{sin}\:\frac{\boldsymbol{\theta}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{xsin}\:\frac{\pi}{\mathrm{7}}=\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2sin}\:\frac{\pi}{\mathrm{7}}} \\ $$$$\:\mathrm{du}\:\mathrm{rayon}\:\mathrm{du}\:\mathrm{cercle}\:\mathrm{circonscrit}\:\boldsymbol{\mathrm{r}}=\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{surface}}\:\boldsymbol{\mathrm{s}}=\boldsymbol{\mathrm{y}}.\boldsymbol{\mathrm{x}}\:\:\:;\:\:\:\:\:\:\:\:\left[\:\boldsymbol{\mathrm{y}}^{\mathrm{2}}…
Question Number 191275 by Mingma last updated on 22/Apr/23 Answered by mr W last updated on 22/Apr/23 Commented by mr W last updated on 22/Apr/23…
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Question Number 191149 by SLVR last updated on 19/Apr/23 $${The}\:{number}\:{of}\:{solutions}\:{of} \\ $$$$\mid{x}+\mathrm{3}\mid+\mid{x}−\mathrm{2}\mid+\mathrm{3}{sinx}−\mathrm{3}=\mathrm{0}\:{is} \\ $$ Commented by SLVR last updated on 19/Apr/23 $${sir}\:{kindly}\:{help}\:{me}\: \\ $$ Commented…
Question Number 191144 by mathlove last updated on 19/Apr/23 Answered by witcher3 last updated on 20/Apr/23 $$\mathrm{tan}\left(\mathrm{a}\right)\mathrm{tan}\left(\mathrm{b}\right)=\frac{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)−\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)+\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)}. \\ $$$$\mathrm{tan}\left(\mathrm{42}\right)\mathrm{tan}\left(\mathrm{78}\right)=\frac{\mathrm{cos}\left(\mathrm{36}\right)+\mathrm{cos}\left(\mathrm{120}\right)}{\mathrm{cos}\left(\mathrm{36}\right)−\mathrm{cos}\left(\mathrm{120}\right)}=\frac{\mathrm{2cos}\left(\mathrm{36}\right)−\mathrm{1}}{\mathrm{2cos}\left(\mathrm{36}\right)+\mathrm{1}} \\ $$$$\mathrm{tan}\left(\mathrm{6}\right)\mathrm{tan}\left(\mathrm{66}\right)=\frac{\mathrm{cos}\left(\mathrm{60}\right)−\mathrm{cos}\left(\mathrm{72}\right)}{\mathrm{cos}\left(\mathrm{60}\right)+\mathrm{cos}\left(\mathrm{72}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{2cos}\left(\mathrm{72}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{72}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{1}−\mathrm{2cos}\left(\mathrm{72}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{72}\right)}.\frac{\mathrm{2cos}\left(\mathrm{36}\right)−\mathrm{1}}{\mathrm{2cos}\left(\mathrm{36}\right)+\mathrm{1}}=\mathrm{1}…
Question Number 191102 by Mingma last updated on 18/Apr/23 Answered by mahdipoor last updated on 18/Apr/23 $${AB}={AC}={a} \\ $$$${AM}={BC}={b}=\sqrt{{a}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {cos}\mathrm{20}}=\mathrm{2}{asin}\mathrm{10} \\ $$$${BM}^{\mathrm{2}} ={AM}^{\mathrm{2}}…
Question Number 190993 by mathlove last updated on 16/Apr/23 $${sin}^{\mathrm{2}} {x}\:\centerdot{cos}^{\mathrm{2}} {x}=? \\ $$ Answered by cortano12 last updated on 16/Apr/23 $$\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x} \\ $$…