Category: Geometry

Question Number 219425 by Nicholas666 last updated on 24/Apr/25 Commented by Nicholas666 last updated on 24/Apr/25 $$\:\:\:\:{ABC}\:{reguler}\:{pentagon}\: \\ $$$$\:\:\:{P},\:{Q},\:{T}\:\:\:\:\:{toricelli}'{s}\:{point}\:{of}\:{AED},{BCD},{ADE}\:\:\:\:\: \\ $$$$\:\:\:{find}\:{angle}\:{PTQ}? \\ $$ Answered by…

Question Number 219337 by Spillover last updated on 23/Apr/25 Answered by mr W last updated on 23/Apr/25 Commented by mr W last updated on 23/Apr/25…

Question Number 219339 by Spillover last updated on 23/Apr/25 Answered by mr W last updated on 23/Apr/25 $$\alpha=\mathrm{45}°+\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\alpha=\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}}{\mathrm{1}−\mathrm{1}×\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{3} \\ $$ Answered by…

Question Number 219293 by Nicholas666 last updated on 22/Apr/25 Commented by Nicholas666 last updated on 22/Apr/25 $${find}\:{the}\:{angle}\:{x}? \\ $$$$ \\ $$ Commented by Ghisom last…

Question Number 219255 by Spillover last updated on 21/Apr/25 Answered by A5T last updated on 21/Apr/25 $$\mathrm{WLOG},\:\mathrm{let}\:\mathrm{the}\:\mathrm{side},\mathrm{s},\:\mathrm{of}\:\mathrm{the}\:\mathrm{square}\:\mathrm{be}\:\mathrm{1} \\ $$$$\Rightarrow\mathrm{b}^{\mathrm{2}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} =\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{a}×\mathrm{b}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{4}}\overset{/\mathrm{b}^{\mathrm{2}} }…