Category: Geometry

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 $$\alpha =45°+{\tan }^{-1}\frac{1}{2}$$$$\tan \alpha =\frac{1+\frac{1}{2}}{1-1\times \frac{1}{2}}=3$$ Answered by…

Question Number 219293 by Nicholas666 last updated on 22/Apr/25 Commented by Nicholas666 last updated on 22/Apr/25 $$findtheanglex?$$$$$$ Commented by Ghisom last…

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

Question Number 219232 by Spillover last updated on 20/Apr/25 Answered by A5T last updated on 21/Apr/25 Commented by A5T last updated on 21/Apr/25 $$\mathrm{AD}=\mathrm{bsin}\theta\:;\:\mathrm{AE}=\mathrm{bcos}\theta\:;\:\mathrm{AC}=\mathrm{asin}\theta\:;\:\mathrm{AB}=\mathrm{acos}\theta \…