Question Number 225885 by Mingma last updated on 15/Nov/25 Commented by Mingma last updated on 15/Nov/25 Can you show workings, Prof? Commented by fantastic2 last updated on 15/Nov/25 $$\angle{DCH}+\angle{DBH}=\mathrm{45}^{\mathrm{0}}…
Question Number 225892 by mnjuly1970 last updated on 15/Nov/25 Answered by A5T last updated on 15/Nov/25 $$\mathrm{Let}\:\mathrm{O}\:\mathrm{be}\:\mathrm{the}\:\mathrm{centre}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{and}\:\mathrm{OX}=\mathrm{d} \\ $$$$\mathrm{Let}\:\mathrm{OB}=\mathrm{r}\Rightarrow\mathrm{XB}=\mathrm{r}−\mathrm{d};\:\mathrm{AX}=\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{e} \\ $$$$\mathrm{AO}=\sqrt{\mathrm{e}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} }\: \\ $$$$\angle\mathrm{BYC}=\mathrm{90}°\:\Rightarrow\:\mathrm{XHYC}\:\mathrm{is}\:\mathrm{cyclic}…
Question Number 225904 by ajfour last updated on 15/Nov/25 Commented by ajfour last updated on 15/Nov/25 $${If}\:{parabola}\:{is}\:{y}={x}^{\mathrm{2}} .\:{Find}\:{equations} \\ $$$${of}\:{both}\:{these}\:{circles}\:{of}\:{equal}\:{radii}. \\ $$ Answered by mr…
Question Number 225856 by ajfour last updated on 15/Nov/25 Commented by ajfour last updated on 15/Nov/25 $${Parabola}\:{shown}\:{is}\:\:{y}={x}^{\mathrm{2}} . \\ $$$${Find}\:{equations}\:{of}\:{maximum} \\ $$$${radius}\:{yellow}\:{circles},\:{or}\:{find} \\ $$$$\:\:\:\:\:\:\:{r},\:{C}\left({h},{k}\right)\:,\:{C}\:'\left(−{h},{k}\right) \\…
Question Number 225756 by ajfour last updated on 09/Nov/25 Commented by mr W last updated on 11/Nov/25 $${applying}\:{descartes}\:{theorem}\:{twice}: \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{r}}}=\frac{\mathrm{1}}{\:\sqrt{{R}}}+\frac{\mathrm{1}}{\:\sqrt{{R}}}=\frac{\mathrm{2}}{\:\sqrt{{R}}}\:\Rightarrow{r}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{{a}}}=\frac{\mathrm{1}}{\:\sqrt{{r}}}+\frac{\mathrm{1}}{\:\sqrt{{R}}}=\frac{\mathrm{3}}{\:\sqrt{{R}}} \\ $$$$\Rightarrow{a}=\frac{{R}}{\mathrm{9}} \\…
Question Number 225703 by ajfour last updated on 07/Nov/25 Answered by mahdipoor last updated on 07/Nov/25 $$\mathrm{2}×\left(\mathrm{R}−\mathrm{r}\right)^{\mathrm{2}} =\left(\mathrm{R}+\mathrm{r}\right)^{\mathrm{2}} \Rightarrow\mathrm{r}=\mathrm{R}\left(\mathrm{3}−\sqrt{\mathrm{8}}\right) \\ $$$$\mathrm{center}\:\mathrm{of}\:\mathrm{balls}\:\left(\mathrm{r}\right)\:\mathrm{in}\:\mathrm{R}_{\mathrm{c}} =\mathrm{R}−\mathrm{r}\:\: \\ $$$$\theta_{\mathrm{each}\:\mathrm{ball}} =\mathrm{2tan}^{−\mathrm{1}}…
Question Number 225656 by Jubr last updated on 05/Nov/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 225657 by Jubr last updated on 05/Nov/25 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 225658 by Jubr last updated on 05/Nov/25 Answered by ajfour last updated on 06/Nov/25 $${p}={R}\mathrm{sin}\:\alpha \\ $$$$\pi−\alpha=\frac{\mathrm{5}}{\mathrm{2}}\:\:\Rightarrow\:\:\mathrm{2}\alpha=\mathrm{2}\pi−\mathrm{5} \\ $$$$\angle{AOC}=\mathrm{2}\left(\frac{\pi}{\mathrm{2}}−\alpha\right)=\pi−\mathrm{2}\alpha \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\pi−\left(\mathrm{2}\pi−\mathrm{5}\right)=\mathrm{5}−\pi \\ $$$${A}_{{shade}}…
Question Number 225666 by mr W last updated on 05/Nov/25 Commented by mr W last updated on 05/Nov/25 $${an}\:{alternative}\:{solution} \\ $$ Commented by ajfour last…