Question Number 190732 by zinaung last updated on 10/Apr/23 Answered by JDamian last updated on 10/Apr/23 $${is}\:\:\sqrt{−\sqrt{\mathrm{3}+{i}}}\:\:\:\:{or}\:\:\sqrt{−\sqrt{\mathrm{3}}+{i}\:\:\:}? \\ $$ Answered by Frix last updated on…
Question Number 125073 by amns last updated on 08/Dec/20 $${If}\:\sqrt{\mathrm{2}}{cos}\left({A}\:−\:{B}\right)\:=\:\mathrm{1},\:\mathrm{2}{sin}\left({A}\:+\:{B}\right)\:=\:\sqrt{\mathrm{3}}\:{and}\:{A},\:{B}\:{is}\:{a}\:{fine} \\ $$$${angle},\:{then}\:{determine}\:{A}'{s}\:{and}\:{B}'{s}\:{value}. \\ $$$$\boldsymbol{{Please}}\:\boldsymbol{{help}}\:\boldsymbol{{me}}. \\ $$ Answered by som(math1967) last updated on 08/Dec/20 $$\mathrm{cos}\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}=\mathrm{cos45} \\…
Question Number 190527 by Rupesh123 last updated on 04/Apr/23 Commented by Frix last updated on 05/Apr/23 $$\pm\frac{\mathrm{5}}{\mathrm{6}} \\ $$ Answered by cortano12 last updated on…
Question Number 59441 by rahul 19 last updated on 10/May/19 Answered by tanmay last updated on 10/May/19 $${sinx}+\mathrm{2}{costcosx}=\mathrm{2} \\ $$$$\mathrm{4}{cos}^{\mathrm{2}} {t}\left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)=\mathrm{4}−\mathrm{4}{sinx}+{sin}^{\mathrm{2}} {x} \\ $$$${cost}={c}\:\:\:{sinx}={k}…
Question Number 124927 by bemath last updated on 07/Dec/20 Answered by liberty last updated on 07/Dec/20 $$\left(\mathrm{1}\right)\:\mathrm{sin}\:\mathrm{38}°\:=\:\frac{{x}+{y}}{\mathrm{8}}\:;\:{x}+{y}\:=\:\mathrm{8}.\mathrm{sin}\:\mathrm{38}° \\ $$$$\Rightarrow{x}+{y}\:=\:\mathrm{4}.\mathrm{9}\:{ft}\: \\ $$$$\left(\mathrm{2}\right)\:\mathrm{sin}\:\mathrm{52}°\:=\:\frac{{t}}{\mathrm{8}}\:;\:{t}\:=\:\mathrm{8}.\mathrm{sin}\:\mathrm{52}°\:=\:\mathrm{6}.\mathrm{3}\:{ft} \\ $$$$\left(\mathrm{3}\right)\:\mathrm{tan}\:\mathrm{20}°\:=\:\frac{{x}}{{t}}\:;\:{x}\:=\:{t}.\mathrm{tan}\:\mathrm{20}°\:=\:\mathrm{2}.\mathrm{3} \\ $$$${then}\:{we}\:{get}\:{y}\:=\:\mathrm{4}.\mathrm{9}−\mathrm{2}.\mathrm{3}\:=\:\mathrm{2}.\mathrm{6}\:{ft}…
Question Number 59326 by prattushdas7@gmail.com last updated on 08/May/19 $${sin}\:{cos}^{−\mathrm{1}} \left(−\sqrt{\left.\mathrm{3}/\mathrm{2}\right)}\right. \\ $$ Commented by MJS last updated on 08/May/19 $$\mathrm{sin}\:\mathrm{cos}^{−\mathrm{1}} \:{x}\:=\mathrm{cos}\:\mathrm{sin}^{−\mathrm{1}} \:{x}\:=\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\…
Question Number 59305 by pete last updated on 07/May/19 $$\mathrm{Witthout}\:\mathrm{using}\:\mathrm{mathematical}\:\mathrm{tables}\:\mathrm{or} \\ $$$$\mathrm{calculator},\:\mathrm{find},\mathrm{in}\:\mathrm{surdform}\left(\mathrm{radicals}\right) \\ $$$$,\:\mathrm{the}\:\mathrm{value}\:\mathrm{tan22}.\mathrm{5}° \\ $$ Answered by ajfour last updated on 07/May/19 $$\mathrm{tan}\:\mathrm{2}\theta=\frac{\mathrm{2tan}\:\theta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \theta}…
Question Number 59246 by hovea cw last updated on 06/May/19 Commented by MJS last updated on 06/May/19 $$\left(\mathrm{1}\right)\:\mathrm{doesn}'\mathrm{t}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{have}\:\mathrm{a}\:“\mathrm{nice}''\:\mathrm{solution} \\ $$$$\mathrm{I}\:\mathrm{get}\:\sqrt{\mathrm{3}}\mathrm{tan}\:\mathrm{10}°\:\mathrm{cos}\:\mathrm{40}°\:\approx.\mathrm{233956} \\ $$ Commented by maxmathsup…
Question Number 190306 by cortano12 last updated on 31/Mar/23 Answered by Frix last updated on 31/Mar/23 $$\mathrm{Area}\:\mathrm{of}\:\mathrm{triangle}\:{u},\:{v},\:{w}\:\mathrm{is} \\ $$$$\frac{\sqrt{\left({u}+{v}+{w}\right)\left({u}+{v}−{w}\right)\left({u}+{w}−{v}\right)\left({v}+{w}−{u}\right)}}{\mathrm{4}}= \\ $$$$=\frac{\sqrt{\mathrm{2}\left({u}^{\mathrm{2}} {v}^{\mathrm{2}} +{u}^{\mathrm{2}} {w}^{\mathrm{2}} +{v}^{\mathrm{2}}…
Question Number 190249 by alcohol last updated on 30/Mar/23 $${if}\:\mathrm{8\%}\:{error}\:{is}\:{made}\:{on}\:{x},\: \\ $$$${what}\:{is}\:{the}\:{percentage}\:{error} \\ $$$${on}\:\lambda{x}^{\frac{\mathrm{1}}{\mathrm{5}}} \:? \\ $$ Answered by mr W last updated on 30/Mar/23…