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…
Question Number 167008 by mnjuly1970 last updated on 04/Mar/22 Answered by mr W last updated on 06/Mar/22 Commented by Tawa11 last updated on 06/Mar/22 $$\mathrm{Great}\:\mathrm{sir}…
Question Number 167011 by mnjuly1970 last updated on 04/Mar/22 Answered by TheSupreme last updated on 04/Mar/22 $${Talete}\:{theorem}\:{respect}\:{point}\:{A}\:{far}\:{away}\:{x}\:{from}\:{center}\: \\ $$$${of}\:{circle}\:{of}\:{radius}\:\mathrm{4} \\ $$$${let}'{s}\:{y}={PQ} \\ $$$${x}:\mathrm{4}=\left({x}+\mathrm{4}\right):{y}=\left({x}+\mathrm{12}\right):\mathrm{8} \\ $$$$\frac{{x}}{\mathrm{4}}=\frac{{x}+\mathrm{12}}{\mathrm{8}}\rightarrow{x}=\mathrm{12}…
Question Number 35871 by MJS last updated on 25/May/18 $$\mathrm{new}\:\mathrm{idea}\:\left(\mathrm{and}\:\mathrm{solution}\right)\:\mathrm{to}\:\mathrm{questions}\:\mathrm{35178}\:\&\:\:\mathrm{35195} \\ $$$$ \\ $$$$\mathrm{triangle}: \\ $$$${ABC};\:{a}={BC},\:{b}={CA},\:{c}={AB}; \\ $$$$\alpha=\angle{CAB},\:\beta=\angle{ABC},\:\gamma=\angle{BCA} \\ $$$${d}=\sqrt{\left({a}+{b}+{c}\right)\left({a}+{b}−{c}\right)\left({a}−{b}+{c}\right)\left(−{a}+{b}+{c}\right)} \\ $$$$ \\ $$$$\mathrm{put}\:\mathrm{it}\:\mathrm{as}\:\mathrm{this}: \\…
Question Number 166928 by amin96 last updated on 02/Mar/22 Answered by mr W last updated on 02/Mar/22 $${a}\:{picture}\:{says}\:{more}\:{than}\:{a}\:{thousant} \\ $$$${words}… \\ $$ Commented by mr…
Question Number 166917 by ajfour last updated on 02/Mar/22 Commented by mr W last updated on 02/Mar/22 Commented by ajfour last updated on 02/Mar/22 $$\left\{\left(\mathrm{2}{b}−{c}\right)\mathrm{cos}\:\theta−{b}\right\}^{\mathrm{2}}…