Question Number 109948 by mnjuly1970 last updated on 26/Aug/20 Answered by 1549442205PVT last updated on 26/Aug/20 $$\mathrm{acosA}=\mathrm{a}.\frac{\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} }{\mathrm{2bc}}=\frac{\mathrm{a}^{\mathrm{2}} \left(\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{a}^{\mathrm{2}} \right)}{\mathrm{2abc}}. \\…
Question Number 175476 by ajfour last updated on 31/Aug/22 Commented by ajfour last updated on 31/Aug/22 $${A}\:{cone}\:{has}\:{an}\:{inscribed}\:{cube}\:{resting} \\ $$$${at}\:{its}\:{base}.\:{Atop}\:{the}\:{cube}\:{inscribed} \\ $$$${in}\:{the}\:{cone}\:{is}\:{a}\:{sphere}\:{of}\:{radius}\:{r}. \\ $$$${Find}\:{r}\:{in}\:{terms}\:{of}\:{R},\:{H}\:\left({cone}'{s}\right. \\ $$$$\left.{radius}\:{and}\:{height}\right).…
Question Number 44395 by ajfour last updated on 28/Sep/18 Commented by ajfour last updated on 28/Sep/18 $${Find}\:{radius}\:{of}\:{the}\:{green}\:{circle}, \\ $$$${in}\:{terms}\:{of}\:{R}\:{and}\:{r}. \\ $$ Commented by ajfour last…
Question Number 175449 by cherokeesay last updated on 30/Aug/22 Commented by som(math1967) last updated on 31/Aug/22 $${is}\:{r}=\mathrm{7}−\mathrm{3}\sqrt{\mathrm{5}}\approx\mathrm{0}.\mathrm{29}\:? \\ $$$$\frac{{S}_{\mathrm{2}} }{{S}_{\mathrm{1}} }\:=\frac{\pi\left(.\mathrm{29}\right)^{\mathrm{2}} }{\pi}×\mathrm{100}=\mathrm{8}.\mathrm{41\%}\:? \\ $$ Answered…
Question Number 44365 by ajfour last updated on 27/Sep/18 Commented by ajfour last updated on 27/Sep/18 $${Find}\:{area}\:{of}\:\bigtriangleup{ABC}\:{in}\:{terms}\:{of} \\ $$$$\alpha,\:\beta,\:{and}\:{R}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last…
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Question Number 109674 by I want to learn more last updated on 24/Aug/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{enclose}\:\mathrm{by}\:\mathrm{the}\:\mathrm{curve}\:\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:−\:\:\mathrm{x}^{\mathrm{2}} \:, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{1}\:\:+\:\:\mathrm{3x},\:\:\:\:\mathrm{x}\:\:=\:\:\mathrm{1}. \\ $$ Answered by bobhans last updated on…
Question Number 175199 by cherokeesay last updated on 22/Aug/22 Answered by som(math1967) last updated on 23/Aug/22 Commented by som(math1967) last updated on 23/Aug/22 $${OLBT}\:{rectangle} \\…
Question Number 175188 by ajfour last updated on 22/Aug/22 Answered by cherokeesay last updated on 22/Aug/22 $$\:\:\:\:\:\:\:\:\:\:\:\:\frac{{AD}}{\mathrm{2}}=\mathrm{1}={OA}={OC}={AC}={PC} \\ $$$$\:\:\:\:\frac{{AB}}{{AC}}=\frac{{PQ}}{{PC}}\Leftrightarrow\frac{{AB}}{\mathrm{1}}=\frac{{PQ}}{\mathrm{1}}={sin}\left(\mathrm{30}°\right)=\frac{\mathrm{1}}{\mathrm{2}}\Leftrightarrow \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{AB}={PQ}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$…
Question Number 175176 by nadovic last updated on 21/Aug/22 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{locus}\:\mathrm{of}\: \\ $$$$\mathrm{points}\:\mathrm{equidistant}\:\mathrm{from}\:\mathrm{the}\:\mathrm{point} \\ $$$${A}\left(\mathrm{4},\:−\mathrm{1}\right)\:\mathrm{and}\:\mathrm{the}\:\mathrm{line}\:{x}−{y}+\mathrm{2}=\mathrm{0}. \\ $$ Answered by som(math1967) last updated on 22/Aug/22 $${let}\:{movale}\:{point}\left({h},{k}\right) \\…