Question Number 167532 by mathls last updated on 19/Mar/22 Commented by mathls last updated on 19/Mar/22 $$? \\ $$ Commented by mathls last updated on…
Question Number 36346 by behi83417@gmail.com last updated on 31/May/18 Commented by behi83417@gmail.com last updated on 31/May/18 $${find}\:{r}\:{in}\:{terms}\:{of}\::{r}_{{a}} ,{r}_{{b}} ,{r}_{{c}} \\ $$ Commented by behi83417@gmail.com last…
Question Number 36314 by behi83417@gmail.com last updated on 31/May/18 Commented by behi83417@gmail.com last updated on 31/May/18 $${BD}={median},{BE}={angular}\:{bisector} \\ $$$${EF}\parallel{BD} \\ $$$${BF}=?\:\left({in}\:{term}\:{of}\:{A}\overset{\blacktriangle} {{B}C}\:{sides}\right) \\ $$$${spicial}\:{case}:\:{AB}−{AC}={d}\left(={const}.\right) \\…
Question Number 36270 by ajfour last updated on 30/May/18 Commented by ajfour last updated on 30/May/18 $${Find}\:{maximum}\:{area}\:{of}\:{a}\:{triangle} \\ $$$${when}\:{its}\:{vertices}\:{lie}\:{on}\:{the} \\ $$$${cirumference}\:{of}\:{three}\:{circles}. \\ $$$$\left({As}\:{shown}\:{in}\:{fig}.\:{the}\:{circles}\right. \\ $$$$\left.{touch}\:{each}\:{other}\:\right).…
Question Number 167339 by mnjuly1970 last updated on 13/Mar/22 Commented by Tawa11 last updated on 19/Mar/22 $$\mathrm{Great}\:\mathrm{sir}. \\ $$ Answered by greogoury55 last updated on…
Question Number 167225 by amin96 last updated on 10/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167073 by neinhaltsieger369 last updated on 05/Mar/22 $$\:\boldsymbol{\mathrm{Let}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{triangle}}\:\:\boldsymbol{\mathrm{A}}\left(−\mathrm{3};\:\mathrm{2}\right),\:\:\boldsymbol{\mathrm{B}}\left(\mathrm{8};\:\mathrm{4}\right),\:\:\boldsymbol{\mathrm{C}}\left(\mathrm{4};\:−\mathrm{6}\right),\:\:\boldsymbol{\mathrm{solve}}\underset{\:} {\overset{\:} {:}} \\ $$$$\:\left(\boldsymbol{\mathrm{a}}\overset{\:} {\right)}\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{perimeter}}; \\ $$$$\:\left(\boldsymbol{\mathrm{b}}\overset{\:} {\right)}\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{area}}; \\ $$$$\:\left(\boldsymbol{\mathrm{c}}\overset{\:} {\right)}\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{base}}; \\ $$$$\:\left(\boldsymbol{\mathrm{d}}\overset{\:} {\right)}\:\boldsymbol{\mathrm{find}}\:\:\boldsymbol{\mathrm{the}}\:\:\boldsymbol{\mathrm{height}}; \\…
Question Number 167043 by mnjuly1970 last updated on 05/Mar/22 Answered by mr W last updated on 05/Mar/22 $${say}\:\angle{B}=\beta \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{3}\:\mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{4}−\mathrm{3}\:\mathrm{cos}\:\mathrm{2}\alpha}=\frac{\mathrm{6}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{6}\:\mathrm{sin}^{\mathrm{2}} \:\alpha} \\ $$$$\frac{\mathrm{sin}\:\beta}{\mathrm{6}}=\frac{\mathrm{sin}\:\alpha}{\mathrm{4}}\:\Rightarrow\mathrm{sin}\:\beta=\frac{\mathrm{3}\:\mathrm{sin}\:\alpha}{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{6}\:\mathrm{sin}\:\alpha\:\mathrm{cos}\:\alpha}{\mathrm{1}+\mathrm{6}\:\mathrm{sin}^{\mathrm{2}}…
Question Number 167013 by amin96 last updated on 04/Mar/22 Answered by mr W last updated on 05/Mar/22 Commented by Tawa11 last updated on 06/Mar/22 $$\mathrm{Great}\:\mathrm{sir}…
Question Number 167009 by mnjuly1970 last updated on 04/Mar/22 Answered by mr W last updated on 04/Mar/22 Commented by mr W last updated on 04/Mar/22…