Question Number 220560 by mehdee7396 last updated on 15/May/25 $${sin}\alpha=\mathrm{0}.\mathrm{8}\:\Rightarrow\:\frac{{BE}}{{EF}}=? \\ $$ Answered by mehdee7396 last updated on 15/May/25 Answered by mr W last updated…
Question Number 220511 by mehdee7396 last updated on 14/May/25 $${AB}=\mathrm{2}{CE}\:\:\&\:\:{DE}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\:\:\: \\ $$$${CE}\:\bot{AB}\:\:\:\&\:\:\:{AD}\bot{BC}\:\:\&\:\:{AB}={AC}\:\:\&\:{EF}\:\bot{BC}\: \\ $$$${BF}=? \\ $$$$ \\ $$ Answered by mehdee7396 last updated on 14/May/25…
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Question Number 219839 by Lekhraj last updated on 02/May/25 Answered by golsendro last updated on 03/May/25 $$\:\mathrm{sin}\:\mathrm{2}\theta\:=\:\mathrm{2}\left(\frac{\sqrt{\mathrm{4}−\sqrt{\mathrm{5}}}\:+\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\mathrm{2}}\:\right)\left(\frac{\sqrt{\mathrm{4}−\sqrt{\mathrm{5}}}−\sqrt{\sqrt{\mathrm{5}}−\mathrm{2}}}{\mathrm{2}}\:\right) \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}−\sqrt{\mathrm{5}}\:−\sqrt{\mathrm{5}}\:+\mathrm{2}\:\right) \\ $$$$\:=\:\mathrm{3}−\sqrt{\mathrm{5}}\: \\ $$ Terms of…
Question Number 219279 by Mingma last updated on 22/Apr/25 Answered by som(math1967) last updated on 22/Apr/25 $$\:\frac{\mathrm{2}{sin}\frac{\beta+\alpha}{\mathrm{2}}{cos}\frac{\alpha−\beta}{\mathrm{2}}}{\mathrm{2sin}\:\frac{\alpha+\beta}{\mathrm{2}}{sin}\frac{\alpha−\beta}{\mathrm{2}}}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:{cot}\frac{\alpha−\beta}{\mathrm{2}}={cot}\frac{\pi}{\mathrm{6}} \\ $$$$\Rightarrow\:\:\alpha−\beta=\frac{\pi}{\mathrm{3}} \\ $$$$\:\mathrm{sin}\:\mathrm{3}\alpha+\mathrm{sin}\:\beta \\ $$$$=\mathrm{2sin}\:\frac{\mathrm{3}\left(\alpha+\beta\right)}{\mathrm{2}}×\mathrm{cos}\:\frac{\mathrm{3}\left(\alpha−\beta\right)}{\mathrm{2}}…
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Question Number 217783 by malwan last updated on 21/Mar/25 $$ \\ $$ Commented by malwan last updated on 21/Mar/25 $$ \\ $$ Commented by malwan…
Question Number 217384 by MATHEMATICSAM last updated on 12/Mar/25 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{sin351}°\:=\:−\:\sqrt{\frac{\mathrm{4}\:−\:\sqrt{\mathrm{10}\:+\:\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{8}}\:.} \\ $$ Answered by Sutrisno last updated on 14/Mar/25 $${cos}\mathrm{18}°=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}} \\ $$$${sin}\mathrm{351}°=−{sin}\mathrm{9}° \\ $$$$=−\sqrt{\frac{\mathrm{1}−{cos}\mathrm{18}°}{\mathrm{2}}} \\…
Question Number 217146 by a.lgnaoui last updated on 02/Mar/25 $$\mathrm{determiner}\:\mathrm{le}\:\mathrm{cote}\:\mathrm{du}\:\mathrm{care}\:\boldsymbol{\mathrm{ABCD}} \\ $$$$\mathrm{inscrit}\:\mathrm{dans}\:\mathrm{l}\:\mathrm{elipse}\:\left\{\left(−\mathrm{3},+\mathrm{3}\right):\left(−\mathrm{8},+\mathrm{8}\right)\right\} \\ $$ Commented by a.lgnaoui last updated on 02/Mar/25 Answered by mr W…
Question Number 216733 by josemanuelrios last updated on 17/Feb/25 $$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Comparto}\:\mathrm{otro}\:\mathrm{reto}\:\mathrm{de}\:\mathrm{matematicas}\:\mathrm{que}\: \\ $$$$\mathrm{dice}\:\mathrm{literalmente}: \\…