Question Number 218887 by Spillover last updated on 17/Apr/25 Answered by mr W last updated on 17/Apr/25 Commented by mr W last updated on 17/Apr/25…
Question Number 218886 by Spillover last updated on 17/Apr/25 Answered by mr W last updated on 17/Apr/25 Commented by Spillover last updated on 17/Apr/25 $${perfect}…
Question Number 218812 by Spillover last updated on 15/Apr/25 Answered by A5T last updated on 16/Apr/25 $$\sqrt{\left(\mathrm{R}−\mathrm{x}\right)^{\mathrm{2}} −\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{x}^{\mathrm{2}} −\left(\mathrm{x}−\mathrm{h}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\left(\mathrm{R}−\mathrm{2x}\right)\mathrm{R}=\mathrm{h}\left(\mathrm{2x}−\mathrm{h}\right) \\ $$$$\Rightarrow\mathrm{R}^{\mathrm{2}}…
Question Number 218813 by Spillover last updated on 15/Apr/25 Answered by nikif99 last updated on 15/Apr/25 Commented by Spillover last updated on 16/Apr/25 $${great}\:{work}.{thanks}\: \\…
Question Number 218778 by Spillover last updated on 15/Apr/25 Answered by mr W last updated on 15/Apr/25 Commented by mr W last updated on 17/Apr/25…
Question Number 218676 by Spillover last updated on 14/Apr/25 Answered by mr W last updated on 14/Apr/25 $$\mathrm{cos}\:\theta=\frac{{R}−{r}}{{R}+{r}}=\mathrm{1}−\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\sqrt{\frac{{r}}{{R}+{r}}} \\ $$$${x}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\mathrm{2}{R}\sqrt{\frac{{r}}{{R}+{r}}}\:\:\checkmark \\ $$…
Question Number 218736 by Spillover last updated on 14/Apr/25 Answered by mr W last updated on 15/Apr/25 Commented by mr W last updated on 15/Apr/25…
Question Number 218737 by Spillover last updated on 14/Apr/25 Answered by Spillover last updated on 16/Apr/25 Answered by Spillover last updated on 16/Apr/25 Terms of…
Question Number 218738 by Spillover last updated on 14/Apr/25 Answered by som(math1967) last updated on 15/Apr/25 Commented by som(math1967) last updated on 15/Apr/25 $${area}\bigtriangleup{ABC}=\frac{\mathrm{1}}{\mathrm{2}}×{a}×{b}×{sinA} \\…
Question Number 218739 by Spillover last updated on 14/Apr/25 Answered by A5T last updated on 15/Apr/25 $$\mathrm{AB}=\mathrm{c}\:;\:\mathrm{BC}=\mathrm{a}\:;\:\mathrm{CA}\:=\mathrm{b}\:;\:\mathrm{AF}=\mathrm{h}_{\mathrm{a}} \\ $$$$\mathrm{AF}\bot\mathrm{BC}\:;\:\mathrm{BG}\bot\mathrm{DC}\:;\:\mathrm{DH}\bot\mathrm{BC}\: \\ $$$$\mathrm{S}_{\bigtriangleup\mathrm{ABC}} =\frac{\mathrm{ah}_{\mathrm{a}} }{\mathrm{2}}\Rightarrow\mathrm{h}_{\mathrm{a}} =\frac{\mathrm{2S}_{\bigtriangleup\mathrm{ABC}} }{\mathrm{a}}…