Question Number 38593 by ajfour last updated on 27/Jun/18 Commented by ajfour last updated on 27/Jun/18 $${Find}\:\theta\:{in}\:{terms}\:{of}\:\boldsymbol{{c}}\:. \\ $$ Commented by MrW3 last updated on…
Question Number 104101 by ajfour last updated on 19/Jul/20 Commented by ajfour last updated on 19/Jul/20 $${If}\:{the}\:{outer}\:{triangle}\:{is}\:{equilateral}, \\ $$$${and}\:{the}\:{two}\:{ellipses}\:{of}\:{same}\:{size} \\ $$$${and}\:{shape},\:{find}\:{ratio}\:{of}\:{area}\:{of} \\ $$$${outer}\:{triangle}\:{to}\:{inner}\left({blue}\right) \\ $$$${triangle}.…
Question Number 169572 by mnjuly1970 last updated on 03/May/22 Answered by mr W last updated on 03/May/22 $${EO}={OC}=\mathrm{1},\:{say} \\ $$$$\angle{HOC}=\angle{AOE}=\angle{B} \\ $$$${AE}=\mathrm{1}×\mathrm{tan}\:{B} \\ $$$$\mathrm{tan}\:{A}=\frac{{EC}}{{AE}}=\frac{\mathrm{2}}{\mathrm{1}×\mathrm{tan}\:{B}} \\…
Question Number 103928 by I want to learn more last updated on 18/Jul/20 Answered by Dwaipayan Shikari last updated on 18/Jul/20 $${Total} \\ $$$${distance}\:{travel}\:{in}\:{x}\:{direction}=\frac{\mathrm{31}}{\mathrm{30}}{km} \\…
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Question Number 103774 by dw last updated on 17/Jul/20 Answered by mr W last updated on 17/Jul/20 $$\boldsymbol{{METHOD}}\:\boldsymbol{{I}} \\ $$$$\overset{\rightarrow} {{AB}}=\boldsymbol{{a}} \\ $$$$\overset{\rightarrow} {{BC}}=\boldsymbol{{b}} \\…
Question Number 103751 by hardylanes last updated on 17/Jul/20 $$ \\ $$$${given}\:{that}\:{the}\:{parametric}\:{equation}\:{of}\: \\ $$$${a}\:{curvd}\:{are}\:{x}=\frac{\mathrm{1}}{{t}−\mathrm{1}}\:\:{y}=\frac{\mathrm{1}}{{t}+\mathrm{1}}\:{obtaun}\:{a}\: \\ $$$${cartesian}\:{equation}\:{of}\:{the}\:{curve}.\:{Hemce}\:{find} \\ $$$${an}\:{equqtion}\:{of}\:{tbe}\:{nirmal}\:{to}\:{the}\:{curve}\:{at}\:{the}\: \\ $$$${point}\:{t}=\mathrm{2} \\ $$ Answered by bobhans…
Question Number 169281 by mnjuly1970 last updated on 28/Apr/22 Commented by cortano1 last updated on 28/Apr/22 $$\mathrm{tan}\:\beta=\frac{\mathrm{1}}{\mathrm{5}},\:\mathrm{tan}\:\left(\alpha+\beta\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{tan}\:\alpha\:=\frac{\mathrm{3}}{\mathrm{11}} \\ $$$$\therefore\:\mathrm{tan}\:\left(\mathrm{45}°+\alpha\right).\mathrm{tan}\:\left(\mathrm{45}°−\beta\right)= \\ $$$$\:\frac{\mathrm{1}+\frac{\mathrm{3}}{\mathrm{11}}}{\mathrm{1}−\frac{\mathrm{3}}{\mathrm{11}}}\:.\frac{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{5}}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{5}}}\:=\:\frac{\mathrm{14}}{\mathrm{8}}\:.\frac{\mathrm{2}}{\mathrm{3}}=\frac{\mathrm{7}}{\mathrm{6}} \\ $$…
Question Number 169226 by Tawa11 last updated on 26/Apr/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 169101 by mnjuly1970 last updated on 24/Apr/22 $$ \\ $$$$\:\:\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\::\:\:\:\:{m}_{{b}} ^{\:\mathrm{2}} \:+\:{m}_{{c}} ^{\:\mathrm{2}} =\:\mathrm{5}\:{m}_{{a}} ^{\:\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\::\:\:\:\overset{\:\:\wedge} {{A}}\:=\:\mathrm{90}^{\:°} \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}_{{a}} :\:\:\left(\:{median}\:\right) \\…