Question Number 200785 by cortano12 last updated on 23/Nov/23 Answered by som(math1967) last updated on 24/Nov/23 $$\:\frac{{AB}}{{sin}\mathrm{105}}=\frac{{AT}}{{sin}\mathrm{45}} \\ $$$$\Rightarrow{AB}=\mathrm{24}×{sin}\mathrm{75}×\sqrt{\mathrm{2}} \\ $$$$\:\measuredangle{ATC}=\measuredangle{ACT}=\mathrm{75} \\ $$$$\therefore{AT}={AC}=\mathrm{24}{cm} \\ $$$$\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{AC}×{sin}\mathrm{60}…
Question Number 200773 by cortano12 last updated on 23/Nov/23 Answered by cherokeesay last updated on 23/Nov/23 Answered by Bharathi last updated on 23/Nov/23 $$\:\:\:\mathrm{Suppose}\:\mathrm{square}\:\mathrm{side}\:\mathrm{be}\:{a} \\…
Question Number 200646 by Calculusboy last updated on 21/Nov/23 Answered by Frix last updated on 21/Nov/23 $$\left(\frac{\mathrm{1}+\mathrm{sin}\:\theta\:+\mathrm{i}\:\mathrm{cos}\:\theta}{\mathrm{1}+\mathrm{sin}\:\theta\:−\mathrm{i}\:\mathrm{cos}\:\theta}\right)^{{n}} =\left(\mathrm{sin}\:\theta\:+\mathrm{i}\:\mathrm{cos}\:\theta\right)^{{n}} = \\ $$$$=\left(\mathrm{cos}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)\:+\mathrm{i}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\theta\right)\right)^{{n}} =\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{\mathrm{2}}−\theta\right){n}} = \\ $$$$=\mathrm{cos}\:\left(\left(\frac{\pi}{\mathrm{2}}−\theta\right){n}\right)\:+\mathrm{i}\:\mathrm{sin}\:\left(\left(\frac{\pi}{\mathrm{2}}−\theta\right){n}\right)…
Question Number 200379 by cortano12 last updated on 18/Nov/23 Commented by Frix last updated on 18/Nov/23 $$\mathrm{No}\:\mathrm{exact}\:\mathrm{solutions}\:\mathrm{possible}.\:\mathrm{Transform}\:\mathrm{to} \\ $$$$\begin{cases}{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}\left({x}+\mathrm{1}\right)}{y}+\frac{{x}^{\mathrm{2}} +{x}−\mathrm{20}}{\mathrm{4}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\\{{y}^{\mathrm{2}} +\frac{{x}^{\mathrm{2}} +\mathrm{6}{x}+\mathrm{6}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}{y}+\frac{\mathrm{3}{x}−\mathrm{20}}{\mathrm{2}{x}\left({x}+\mathrm{1}\right)}=\mathrm{0}}\end{cases} \\…
Question Number 200268 by cortano12 last updated on 16/Nov/23 Answered by ajfour last updated on 16/Nov/23 $${D}\:{origin}. \\ $$$${O}_{\mathrm{2}} \equiv\left({s}−{r},\:{s}−{r}\right) \\ $$$${O}_{\mathrm{3}} \equiv\left({r},\:{s}−{r}\right) \\ $$$${let}\:\:{eqn}\:{of}\:{line}\:{DF}\:{be}\:…
Question Number 200040 by mnjuly1970 last updated on 12/Nov/23 $$ \\ $$$$\:\:\:\:\:{Q}:\:\:{If}\:\:,\:\:{tan}\left(\frac{\pi}{\mathrm{4}}\:−\alpha\:\right)=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\Rightarrow{Find}\:{the}\:{value}\:{of}\:,\:{tan}\left(\mathrm{4}\alpha\right)=? \\ $$$$ \\ $$ Answered by witcher3 last updated on 12/Nov/23…
Question Number 199968 by a.lgnaoui last updated on 11/Nov/23 $$\mathrm{determiner}\:\boldsymbol{\mathrm{x}}\:\:? \\ $$ Commented by a.lgnaoui last updated on 11/Nov/23 Answered by mr W last updated…
Question Number 199952 by Abdullahrussell last updated on 11/Nov/23 Answered by cortano12 last updated on 11/Nov/23 $$\:=\:\frac{\mathrm{2sin}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:+\:\mathrm{2sin}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{6}\theta}{\mathrm{2cos}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{2}\theta\:+\:\mathrm{2cos}\:\mathrm{7}\theta\:\mathrm{cos}\:\mathrm{6}\theta} \\ $$$$\:=\:\frac{\mathrm{sin}\:\mathrm{7}\theta\left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{6}\theta\right)}{\mathrm{cos}\:\mathrm{7}\theta\left(\mathrm{cos}\:\mathrm{2}\theta+\mathrm{cos}\:\mathrm{6}\theta\right)} \\ $$$$\:=\:\mathrm{tan}\:\mathrm{7}\theta \\ $$ Terms of…
Question Number 199830 by cortano12 last updated on 10/Nov/23 $$\:\:\mathrm{Si}\:\mathrm{cos}\:\mathrm{x}+\mathrm{sin}\:\mathrm{x}=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\mathrm{2}}\:, \\ $$$$\:\mathrm{halle}\:\mathrm{el}\:\mathrm{valor}\:\mathrm{de}\:\mathrm{la}\:\mathrm{expresion}\: \\ $$$$\:\mathrm{R}=\:\mathrm{16}\left(\mathrm{sin}\:^{\mathrm{6}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{6}} \mathrm{x}\right)+\mathrm{3}\left(\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}+\mathrm{csc}^{\mathrm{2}} \:\mathrm{x}\right) \\ $$ Answered by Frix last updated…
Question Number 199821 by a.lgnaoui last updated on 09/Nov/23 $$\mathrm{ABFE}\:\:\mathrm{Care} \\ $$$$\mathrm{determiner}\:\boldsymbol{\mathrm{x}}\:\mathrm{en}\:\mathrm{fonction}\:\mathrm{de}\:\mathrm{a}\:\mathrm{etb} \\ $$$$\mathrm{BC}=\boldsymbol{\mathrm{a}}\:\:\:\:\:\:\mathrm{DE}=\:\boldsymbol{\mathrm{b}} \\ $$ Commented by a.lgnaoui last updated on 09/Nov/23 Commented by…