Question Number 214220 by Ari last updated on 01/Dec/24 Answered by mr W last updated on 01/Dec/24 $${say}\:{a}>{b} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${a}−{b}=\mathrm{6}\:\:\:…\left({ii}\right)…
Question Number 214199 by mr W last updated on 01/Dec/24 Commented by mr W last updated on 01/Dec/24 $$\left[{see}\:{Q}\mathrm{214176}\right] \\ $$$${find}\:{the}\:{radius}\:{of}\:{the}\:{maximal} \\ $$$${circle}\:{inscribed}\:{between}\:{the}\:{curves} \\ $$$${f}\left({x}\right)\:{and}\:{g}\left({x}\right)\:{as}\:{shown}.…
Question Number 214100 by ajfour last updated on 28/Nov/24 Answered by ajfour last updated on 29/Nov/24 Commented by ajfour last updated on 29/Nov/24 $$\mathrm{sin}\:\alpha=\frac{{a}}{\mathrm{2}{b}+{a}}=\frac{\mathrm{1}}{\mathrm{2}{s}+\mathrm{1}}\:\:\:\forall\:\:{s}=\frac{{b}}{{a}},\:{t}=\frac{{r}}{{a}} \\…
Question Number 214040 by ajfour last updated on 25/Nov/24 Commented by ajfour last updated on 25/Nov/24 $${The}\:\bigtriangleup\:{is}\:{equilateral}.\:{Find}\:{its}\:{side}\:\boldsymbol{{s}}. \\ $$ Answered by mr W last updated…
Question Number 214012 by ajfour last updated on 24/Nov/24 Commented by ajfour last updated on 24/Nov/24 $${Outer}\:{circle}\:{radius}\:{is}\:{R}.\:{Circle}\:{with} \\ $$$${center}\:{A}\:{has}\:{radius}\:{r}={R}/\mathrm{2}. \\ $$$${If}\:\bigtriangleup{ABC}\:{is}\:{equilateral},\:{find}\:{its} \\ $$$${edge}\:{length}\:\left({say}\:{a}\right). \\ $$…
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Question Number 213884 by mr W last updated on 20/Nov/24 Commented by mr W last updated on 20/Nov/24 $${find}\:{maximum}\:{radius}\:{of}\:{circle}\:{C} \\ $$ Commented by Frix last…
Question Number 213888 by mnjuly1970 last updated on 20/Nov/24 Commented by mnjuly1970 last updated on 20/Nov/24 $$\:\:\:\:\:{circle}\:{is}\:{tangant}\:{to}\:{parabola} \\ $$$$\:\:.{A}\:{is}\:{center}\:{of}\:{circle} \\ $$$$\:\:\:\:{Find}\:\:\:\:,\:\:{inf}\:\left(\:{R}\:\right)=? \\ $$ Commented by…
Question Number 213859 by ajfour last updated on 19/Nov/24 Commented by mr W last updated on 19/Nov/24 $$\mathrm{0}<{AB}<\mathrm{1} \\ $$$${no}\:{maximum}\:{or}\:{minimum}\:{exists}. \\ $$ Commented by ajfour…
Question Number 213835 by BaliramKumar last updated on 18/Nov/24 Answered by mehdee7396 last updated on 18/Nov/24 $${OA}=\sqrt{\mathrm{13}}\:\:\&\:\:\:{OB}=\mathrm{3} \\ $$$${sin}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}\Rightarrow{cos}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$$\Rightarrow{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\theta=\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{2}}\:\:\checkmark \\ $$$$ \\…