Question Number 60987 by ajfour last updated on 28/May/19 Commented by ajfour last updated on 28/May/19 $${Find}\:{the}\:{illuminated}\:{area}\:{of} \\ $$$${the}\:{inner}\:{curved}\:{surface}\:{of}\: \\ $$$${shown},\:{hollow}\:{open}\:{cylinder}. \\ $$$$\:\:\:\:\:\:\left[{where}\:\:{a}>\mathrm{2}{R}\left(\frac{{H}}{{h}}−\mathrm{1}\right)\right]\:\:\:\:\: \\ $$…
Question Number 126509 by Algoritm last updated on 21/Dec/20 Answered by mr W last updated on 21/Dec/20 Commented by talminator2856791 last updated on 22/Dec/20 $$\:\mathrm{so}\:\mathrm{then}\:\mathrm{the}\:\mathrm{answer}?…
Question Number 192001 by ajfour last updated on 05/May/23 Commented by ajfour last updated on 05/May/23 $${Find}\:{x},\:{in}\:{terms}\:{of}\:{a},\:{b},\:{m}=\mathrm{tan}\:\theta. \\ $$ Answered by ajfour last updated on…
Question Number 60906 by necx1 last updated on 27/May/19 Commented by tanmay last updated on 27/May/19 $${pls}\:{check}\:{question}\:{no}\:\mathrm{60812} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 191962 by ajfour last updated on 04/May/23 Answered by mr W last updated on 04/May/23 $$\left[\sqrt{\left({a}+{x}\right)^{\mathrm{2}} −\left({a}−{x}\right)^{\mathrm{2}} }−\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }\right]^{\mathrm{2}} =\left({b}+{x}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}−{b}−{x}\right)^{\mathrm{2}} \\…
Question Number 60888 by ajfour last updated on 26/May/19 Commented by ajfour last updated on 27/May/19 $${If}\:{length}\:\boldsymbol{{l}}\:{encloses}\:{maximum} \\ $$$${such}\:{area},\:{find}\:{the}\:{area}. \\ $$$$\left({a}\right)\:{if}\:{l}\:{be}\:{circular} \\ $$$$\left({b}\right)\:{if}\:{shape}\:{of}\:{l}\:{not}\:{be}\:{given} \\ $$$$\left({someone}\:{please}\:{consider}\right.…
Question Number 191873 by cherokeesay last updated on 02/May/23 Answered by mehdee42 last updated on 03/May/23 $${let}\:\::\:\angle{RPS}={p}_{\mathrm{1}} \:\&\angle\:{QPS}={p}_{\mathrm{2}} \\ $$$${p}_{\mathrm{1}} =\mathrm{45}−{x}\:\:\:,\:\:{p}_{\mathrm{2}} =\mathrm{135}−\mathrm{3}{x} \\ $$$$\frac{{PS}}{{QS}}=\frac{{sin}\mathrm{3}{x}}{{sinp}_{\mathrm{2}} }=\frac{{sinx}}{{sinp}_{\mathrm{1}}…
Question Number 191856 by TUN last updated on 02/May/23 Answered by Subhi last updated on 02/May/23 $$ \\ $$$${put}\:{AD}={y} \\ $$$$\frac{{BD}}{{sin}\left(\mathrm{30}\right)}=\frac{{y}}{{sin}\left(\mathrm{10}\right)} \\ $$$${BD}=\frac{{sin}\left(\mathrm{30}\right)}{{sin}\left(\mathrm{10}\right)}{y}\:\:\:\:\:\:\:{by}\left[{sin}\:{law}\right] \\ $$$$\frac{{CD}}{{sin}\left(\mathrm{50}\right)}=\frac{{y}}{{sin}\left({x}\right)}\:.{hence}\:{CD}=\frac{{sin}\left(\mathrm{50}\right)}{{sin}\left({x}\right)}{y}…
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Question Number 191841 by Mingma last updated on 01/May/23 Answered by som(math1967) last updated on 01/May/23 $${from}\bigtriangleup{ADC} \\ $$$$\:\frac{{AD}}{{sin}\mathrm{30}}=\frac{{AC}}{{sin}\mathrm{110}} \\ $$$$\Rightarrow\frac{{AD}}{\mathrm{sin}\:\mathrm{30}}=\frac{{AC}}{\mathrm{sin}\:\mathrm{70}}\:……\mathrm{1} \\ $$$${from}\:\bigtriangleup{ABD} \\ $$$$\:\frac{\mathrm{sin}\:\mathrm{20}}{{AD}}=\frac{\mathrm{sin}\:\mathrm{40}}{{BD}}\:\:\:…..\mathrm{2}…