Question Number 125654 by ajfour last updated on 12/Dec/20 Commented by ajfour last updated on 12/Dec/20 $${Find}\:{R}/{r}. \\ $$ Answered by mr W last updated…
Question Number 191191 by mathlove last updated on 20/Apr/23 Answered by mehdee42 last updated on 20/Apr/23 $$\mathrm{14} \\ $$ Answered by mr W last updated…
Question Number 191169 by Mingma last updated on 19/Apr/23 Answered by mehdee42 last updated on 19/Apr/23 $${c}:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow{m}_{{l}} ={y}'_{{A}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow\:{l}:\:\:{y}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left({x}−\mathrm{4}\right)\: \\ $$$${m}_{{d}} ×{m}_{{l}} =−\mathrm{1}\Rightarrow{m}_{{l}}…
Question Number 125632 by Mathgreat last updated on 12/Dec/20 Answered by snipers237 last updated on 13/Dec/20 $${CosA}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{bc}}\:\Rightarrow\:{sin}^{\mathrm{2}} \left(\frac{{A}}{\mathrm{2}}\right)=\frac{\mathrm{1}−{cosA}}{\mathrm{2}}=\frac{{a}^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} }{\mathrm{4}{bc}}<\frac{{a}^{\mathrm{2}} }{\mathrm{4}{bc}} \\…
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Question Number 60056 by bhanukumarb2@gmail.com last updated on 17/May/19 Commented by maxmathsup by imad last updated on 17/May/19 $${we}\:{see}\:{that}\:\:{S}\:=\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{sin}\left({k}\theta\right){cos}^{{k}} \theta}{{k}!}\:\Rightarrow{S}\:={Im}\left(\sum_{{k}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{ik}\theta} \:{cos}^{{k}}…
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