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 $${OA}=\sqrt{\mathrm{13}}\:\:\&\:\:\:{OB}=\mathrm{3} \\ $$$${sin}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}}\Rightarrow{cos}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$$\Rightarrow{tan}\frac{\theta}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}}\Rightarrow\theta=\mathrm{2}{tan}^{−\mathrm{1}} \frac{\mathrm{3}}{\mathrm{2}}\:\:\checkmark \\ $$$$ \\…
Question Number 213818 by ajfour last updated on 17/Nov/24 Commented by ajfour last updated on 17/Nov/24 $${Find}\:{R}\:{in}\:{terms}\:{smaller}\:{radii}\:{a},\:{b}. \\ $$ Answered by mr W last updated…
Question Number 213668 by mr W last updated on 13/Nov/24 Commented by mr W last updated on 13/Nov/24 $${find}\:{shaded}\:{area} \\ $$ Answered by BHOOPENDRA last…
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Question Number 213650 by MathematicalUser2357 last updated on 12/Nov/24 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{pythagorean}\:\mathrm{theorem}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} ={c}^{\mathrm{2}} \:\mathrm{exist}. \\ $$ Answered by MathematicalUser2357 last updated on 12/Nov/24 $$ \\ $$$$…