Question Number 97649 by bagjamath last updated on 09/Jun/20 Commented by bagjamath last updated on 09/Jun/20 $${please},\:{share}\:{the}\:{solution} \\ $$ Commented by bagjamath last updated on…
Question Number 163148 by ajfour last updated on 04/Jan/22 Commented by mr W last updated on 05/Jan/22 Commented by mr W last updated on 05/Jan/22…
Question Number 163143 by amin96 last updated on 04/Jan/22 Commented by HongKing last updated on 04/Jan/22 $$\mathrm{r}^{\mathrm{2}} \:=\:\mathrm{r}\:+\:\mathrm{2021}\:+\:\mathrm{2021}^{\mathrm{2}} \\ $$$$\mathrm{r}^{\mathrm{2}} \:-\:\mathrm{r}\:-\:\mathrm{2021}^{\mathrm{2}} \:-\:\mathrm{2021}\:=\:\mathrm{0} \\ $$$$\mathrm{r}\:=\:-\:\mathrm{2021}\:;\:\mathrm{2022}\:\Rightarrow\:\mathrm{r}\:=\:\mathrm{2022}\:\checkmark \\…
Question Number 97557 by Power last updated on 08/Jun/20 Answered by mr W last updated on 08/Jun/20 $$\angle{CBE}=\angle{EBD}=\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$$\Rightarrow\mathrm{2}\theta=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$${BD}=\mathrm{2}{r}\:\mathrm{sin}\:\alpha \\ $$$${EB}=\frac{\mathrm{2}{r}}{\mathrm{cos}\:\theta} \\…
Question Number 97371 by shaxzod last updated on 07/Jun/20 Commented by mr W last updated on 07/Jun/20 $${see}\:{Q}\mathrm{97033} \\ $$ Terms of Service Privacy Policy…
Question Number 97192 by bagjamath last updated on 07/Jun/20 Commented by bagjamath last updated on 07/Jun/20 $${a}+{b}+{c}+{d}+{e}=… \\ $$ Commented by prakash jain last updated…
Question Number 162598 by amin96 last updated on 30/Dec/21 Answered by mr W last updated on 30/Dec/21 Commented by mr W last updated on 30/Dec/21…
Question Number 162530 by mr W last updated on 30/Dec/21 Commented by mr W last updated on 30/Dec/21 $${find}\:{the}\:{sum}\:{of}\:{areas}\:{of}\:{all}\:\left({infinite}\right) \\ $$$${red}\:{circles} \\ $$ Commented by…
Question Number 162490 by amin96 last updated on 29/Dec/21 Answered by mr W last updated on 30/Dec/21 Commented by mr W last updated on 30/Dec/21…
Question Number 162481 by mnjuly1970 last updated on 29/Dec/21 Answered by mr W last updated on 29/Dec/21 $${length}\:{of}\:{base}\:={h} \\ $$$$\mathrm{tan}\:\alpha=\frac{{b}}{{h}} \\ $$$$\mathrm{tan}\:\mathrm{2}\alpha=\frac{{b}+{a}}{{h}} \\ $$$$\mathrm{tan}\:\mathrm{3}\alpha=\frac{{b}+{a}+{x}}{{h}} \\…