Question Number 49148 by mr W last updated on 03/Dec/18 Commented by mr W last updated on 04/Dec/18 $${This}\:{is}\:{not}\:{a}\:{question},\:{but}\:{an}\:{useful} \\ $$$${information}. \\ $$$$ \\ $$$${Maybe}\:{someone}\:{has}\:{delight}\:{to}\:{prove}\:{it}.…
Question Number 180191 by cherokeesay last updated on 08/Nov/22 $$ \\ $$$$\mathrm{A}\:\mathrm{solid}\:\mathrm{cone}\:\mathrm{of}\:\mathrm{height}\:\mathrm{56cm}\:\mathrm{and}\:\mathrm{a}\:\mathrm{basem} \\ $$$$\mathrm{diaeter}\:\mathrm{of}\:\mathrm{56cm}\:\mathrm{is}\:\mathrm{formed}\:\mathrm{from}\:\mathrm{al} \\ $$$$\mathrm{cylindrica}\:\mathrm{drum}\:\mathrm{equal}\:\mathrm{in}\:\mathrm{height}\:\mathrm{andt} \\ $$$$\mathrm{diameer}\:\mathrm{with}\:\mathrm{the}\:\mathrm{cone}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{surfacee} \\ $$$$\mathrm{ara}\:\mathrm{of}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{drum}. \\ $$$$ \\ $$$$\left(\mathrm{if}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{accompanied}\:\mathrm{by}\:\mathrm{a}\:\right. \\…
Question Number 49060 by Pk1167156@gmail.com last updated on 02/Dec/18 Answered by mr W last updated on 02/Dec/18 $${way}\:\mathrm{1}: \\ $$$$\frac{\mathrm{3}}{\mathrm{sin}\:{C}}=\frac{\mathrm{7}}{\mathrm{sin}\:{A}}=\frac{{x}}{\mathrm{sin}\:{B}}=\frac{{x}}{\mathrm{sin}\:\left({A}+{C}\right)} \\ $$$$\mathrm{sin}\:{C}=\frac{\mathrm{3}\:\mathrm{sin}\:{A}}{\mathrm{7}}=\frac{\mathrm{3}\:\mathrm{sin}\:\mathrm{120}°}{\mathrm{7}}=\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{14}} \\ $$$$\mathrm{cos}\:{C}=\frac{\sqrt{\mathrm{14}^{\mathrm{2}} −\mathrm{9}×\mathrm{3}}}{\mathrm{14}}=\frac{\mathrm{13}}{\mathrm{14}}…
Question Number 180126 by mr W last updated on 07/Nov/22 Commented by HeferH last updated on 07/Nov/22 Answered by HeferH last updated on 07/Nov/22 $${x}^{\mathrm{2}}…
Question Number 180104 by mr W last updated on 07/Nov/22 Commented by mr W last updated on 07/Nov/22 $${A}\:{lies}\:{on}\:{x}−{axis}\:{and}\:{B}\:{on}\:{the}\:{curve} \\ $$$${y}={x}^{\mathrm{2}} .\:{P}\:{is}\:{at}\:\left(\mathrm{4},\mathrm{4}\right).\:{find}\:{the}\:{smallest} \\ $$$${perimeter}\:{of}\:{triangle}\:{PAB}. \\…
Question Number 180066 by mr W last updated on 06/Nov/22 Commented by mr W last updated on 06/Nov/22 $${A}\:{lies}\:{on}\:{the}\:{x}\:{axis}\:{and}\:{B}\:{on}\:{the}\:{blue} \\ $$$${line}.\:{P}\:{is}\:{at}\:\left(\mathrm{4},\mathrm{4}\right).\:{find}\:{the}\:{smallest}\: \\ $$$${perimeter}\:{of}\:{triangle}\:{PAB}. \\ $$…
Question Number 180043 by mr W last updated on 06/Nov/22 Commented by CElcedricjunior last updated on 06/Nov/22 $$\boldsymbol{{d}}'\boldsymbol{{apres}}\:\boldsymbol{{le}}\:\boldsymbol{{theoreme}}\:\boldsymbol{{des}}\:\boldsymbol{{sinus}} \\ $$$$\frac{\mathrm{7}}{\boldsymbol{{sin}\alpha}}=\frac{?}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\alpha}}=\frac{\mathrm{8}}{\boldsymbol{{sin}}\left(\boldsymbol{\pi}−\mathrm{3}\boldsymbol{\alpha}\right)} \\ $$$$=>\frac{\mathrm{7}}{\boldsymbol{{sin}\alpha}}=\frac{?}{\boldsymbol{{sin}}\mathrm{2}\boldsymbol{\alpha}}=\frac{\mathrm{8}}{\boldsymbol{{sin}}\mathrm{3}\boldsymbol{\alpha}} \\ $$$$\Leftrightarrow\frac{?}{\mathrm{7}}=\mathrm{2}\boldsymbol{{cos}\alpha}=>?=\mathrm{14}\boldsymbol{{cos}\alpha}=>?^{\mathrm{2}} =\mathrm{14}^{\mathrm{2}}…
Question Number 48879 by behi83417@gmail.com last updated on 29/Nov/18 Commented by behi83417@gmail.com last updated on 29/Nov/18 $${ABCD}:{square}\:{with}\:{side}\:{length}=\boldsymbol{\mathrm{a}} \\ $$$$\boldsymbol{\mathrm{let}}:\begin{cases}{}&{\boldsymbol{\mathrm{p}}={EF}+{FG}+{GH}+{HE}}\\{}&{\boldsymbol{\mathrm{q}}={EF}^{\mathrm{2}} +{FG}^{\mathrm{2}} +{GH}^{\mathrm{2}} +{HE}^{\mathrm{2}} }\end{cases} \\ $$$${find}:{max}\:\&{min}\:{of}:\:\boldsymbol{\mathrm{p\&q}}.…
Question Number 48874 by behi83417@gmail.com last updated on 29/Nov/18 Commented by behi83417@gmail.com last updated on 29/Nov/18 $${ABC},{equilateral}. \\ $$$${CD}={CE}=\mathrm{8},{AD}={EB}=\mathrm{4},{DE}=\mathrm{3} \\ $$$$\Rightarrow\:\:{AB}=? \\ $$ Answered by…
Question Number 179938 by Acem last updated on 04/Nov/22 Commented by Acem last updated on 04/Nov/22 $${It}'{s}\:{correct},\:{Thanks}\:{Sir}! \\ $$ Commented by HeferH last updated on…