Question Number 191144 by mathlove last updated on 19/Apr/23 Answered by witcher3 last updated on 20/Apr/23 $$\mathrm{tan}\left(\mathrm{a}\right)\mathrm{tan}\left(\mathrm{b}\right)=\frac{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)−\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)}{\mathrm{cos}\left(\mathrm{a}−\mathrm{b}\right)+\mathrm{cos}\left(\mathrm{a}+\mathrm{b}\right)}. \\ $$$$\mathrm{tan}\left(\mathrm{42}\right)\mathrm{tan}\left(\mathrm{78}\right)=\frac{\mathrm{cos}\left(\mathrm{36}\right)+\mathrm{cos}\left(\mathrm{120}\right)}{\mathrm{cos}\left(\mathrm{36}\right)−\mathrm{cos}\left(\mathrm{120}\right)}=\frac{\mathrm{2cos}\left(\mathrm{36}\right)−\mathrm{1}}{\mathrm{2cos}\left(\mathrm{36}\right)+\mathrm{1}} \\ $$$$\mathrm{tan}\left(\mathrm{6}\right)\mathrm{tan}\left(\mathrm{66}\right)=\frac{\mathrm{cos}\left(\mathrm{60}\right)−\mathrm{cos}\left(\mathrm{72}\right)}{\mathrm{cos}\left(\mathrm{60}\right)+\mathrm{cos}\left(\mathrm{72}\right)} \\ $$$$=\frac{\mathrm{1}−\mathrm{2cos}\left(\mathrm{72}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{72}\right)} \\ $$$$\Leftrightarrow\frac{\mathrm{1}−\mathrm{2cos}\left(\mathrm{72}\right)}{\mathrm{1}+\mathrm{2cos}\left(\mathrm{72}\right)}.\frac{\mathrm{2cos}\left(\mathrm{36}\right)−\mathrm{1}}{\mathrm{2cos}\left(\mathrm{36}\right)+\mathrm{1}}=\mathrm{1}…
Question Number 191147 by Shrinava last updated on 19/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191146 by Shrinava last updated on 19/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191143 by Mingma last updated on 19/Apr/23 Answered by HeferH last updated on 21/Apr/23 $$\mathrm{Say}\:\mathrm{CD}\:=\:\mathrm{a},\:\mathrm{DE}\:=\:\mathrm{b},\:\mathrm{AC}\:=\:\mathrm{c},\:\mathrm{AE}\:=\:\mathrm{d} \\ $$$$\:\ast\:\frac{\mathrm{1}}{\mathrm{a}}\:+\:\frac{\mathrm{1}}{\mathrm{b}}\:=\:\frac{\mathrm{1}}{\mathrm{5}}\:\Rightarrow \\ $$$$\:\mathrm{5}\left(\mathrm{b}−\mathrm{a}\right)=\:\mathrm{ab} \\ $$$$\:\ast\:\mathrm{cd}\:=\:\mathrm{ab}+\mathrm{64} \\ $$$$\:\mathrm{Since}\:\mathrm{AD}\:\mathrm{is}\:\mathrm{bisector}:…
Question Number 191142 by Mingma last updated on 19/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191138 by Shrinava last updated on 18/Apr/23 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 191129 by Mastermind last updated on 18/Apr/23 $$\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\:: \\ $$$$\mathrm{x}\:+\:\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\:=\:\mathrm{0}.\mathrm{1614} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$ Answered by Skabetix last updated on 18/Apr/23…
Question Number 191128 by Mingma last updated on 18/Apr/23 Answered by mr W last updated on 18/Apr/23 Commented by mr W last updated on 19/Apr/23…
Question Number 60058 by ajfour last updated on 17/May/19 Commented by ajfour last updated on 17/May/19 $$\mathrm{Find}\:\mathrm{a}\:\mathrm{and}\:\mathrm{h}\:\mathrm{of}\:\mathrm{largest}\:\mathrm{volume}\:\mathrm{prism} \\ $$$$\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{R}. \\ $$ Answered by mr W…
Question Number 191131 by TUN last updated on 18/Apr/23 Answered by a.lgnaoui last updated on 19/Apr/23 $$\bigtriangleup=\left(\mathrm{2m}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{m}^{\mathrm{2}} −\mathrm{6}\right)=\mathrm{25}−\mathrm{4m} \\ $$$$\bigtriangleup>\mathrm{0}\:\:\:\Rightarrow\mathrm{pour}\:\:\:\:\mathrm{m}<\frac{\mathrm{25}}{\mathrm{4}}\:\:\:\:\:\:\mathrm{2}\:\mathrm{solutions} \\ $$$$\mathrm{x}_{\mathrm{1}} =\frac{\mathrm{2m}−\mathrm{1}−\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}};\:\:\mathrm{x}_{\mathrm{2}} =\frac{\mathrm{2m}−\mathrm{1}+\sqrt{\mathrm{25}−\mathrm{4m}}}{\mathrm{2}}…