Question Number 53570 by ajfour last updated on 23/Jan/19 Commented by ajfour last updated on 23/Jan/19 $${Regular}\:{pentagon}\:{side}\:{a}.\:{Find}\:{the} \\ $$$${central}\:{uncoloured}\:{area}. \\ $$ Commented by tanmay.chaudhury50@gmail.com last…
Question Number 184633 by ajfour last updated on 09/Jan/23 Answered by mr W last updated on 09/Jan/23 Commented by mr W last updated on 09/Jan/23…
Question Number 184638 by HeferH last updated on 09/Jan/23 Commented by HeferH last updated on 09/Jan/23 $${find}\:“{x}''\:{if}\:{the}\:{base}\:\left(\mathrm{60}°\:-\:\mathrm{2}\alpha\right)\:{has}\:{a}\:{length}\:{of}\:\mathrm{12} \\ $$ Commented by mr W last updated…
Question Number 184609 by cherokeesay last updated on 09/Jan/23 Answered by ajfour last updated on 09/Jan/23 $$\mathrm{cos}\:\left(\mathrm{2}\alpha−\mathrm{90}°\right)=\frac{\mathrm{2}}{{R}} \\ $$$$\Rightarrow\:\:\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{2}}{{R}}={t} \\ $$$$\mathrm{tan}\:\alpha=\frac{{R}+\sqrt{{R}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}{t}} \\ $$$$\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{2tan}\:\alpha}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}}…
Question Number 184600 by mathlove last updated on 09/Jan/23 Commented by mr W last updated on 09/Jan/23 $${what}\:{is}\:{this}? \\ $$$${what}\:{is}\:{b}? \\ $$$${please}\:{check}\:{if}\:{it}\:{is}\:{really}\:{a}\:{question} \\ $$$${before}\:{posting}\:{it}! \\…
Question Number 53530 by MJS last updated on 23/Jan/19 Commented by MJS last updated on 23/Jan/19 $$\mathrm{I}\:\mathrm{found}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}\:\mathrm{question}\:\mathrm{53168}\:\mathrm{by} \\ $$$$\mathrm{Sir}\:\mathrm{Aifour} \\ $$ Answered by MJS last…
Question Number 184557 by a.lgnaoui last updated on 08/Jan/23 $${Determiner} \\ $$$$\mathrm{1}\bullet\mathrm{AB},\:\:\mathrm{BC}\:\:\mathrm{AC}\:\mathrm{en}\:\mathrm{fonction}\:\mathrm{de}\:\boldsymbol{\mathrm{r}} \\ $$$$\mathrm{2}\bullet\:\:\measuredangle\mathrm{CBA}\:;\:\:\measuredangle\mathrm{BAC}\:;{et}\:\measuredangle\mathrm{BCA} \\ $$ Commented by a.lgnaoui last updated on 08/Jan/23 Answered by…
Question Number 184511 by HeferH last updated on 07/Jan/23 $${In}\:{a}\:{regular}\:{heptagon}\:{ABCDEFG}\:: \\ $$$$\:\sqrt{\mathrm{2}\left({AC}\right)^{\mathrm{2}} −\left({AD}\right)^{\mathrm{2}} }\:−\:{AD}\:=\:\mathrm{2} \\ $$$$\:{find}\:{BC}.\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 53422 by ajfour last updated on 21/Jan/19 Commented by ajfour last updated on 21/Jan/19 $${Find}\:{maximum}\:{inradius}\:{and} \\ $$$${minimum}\:{circumradius}\:{in}\:{terms} \\ $$$${of}\:{a}\:{and}\:{b}\:{if}\:{AB}\:{of}\:\bigtriangleup{ABC}\:{is}\:{variable}. \\ $$ Commented by…
Question Number 184441 by ajfour last updated on 06/Jan/23 Commented by mr W last updated on 07/Jan/23 $${a}=\frac{\sqrt{\mathrm{3}}{Rr}}{\:\sqrt{{R}^{\mathrm{2}} +{r}^{\mathrm{2}} −{Rr}}} \\ $$ Answered by ajfour…