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Category: Geometry

Question-206750

Question Number 206750 by mr W last updated on 23/Apr/24 Answered by HeferH24 last updated on 23/Apr/24 $$\:{By}\:{similar}\:{triangles}: \\ $$$$\:\frac{{d}}{{c}}\:=\:\frac{{x}}{{b}}\:;\:\:\frac{{d}}{{c}}\:=\:\frac{{a}}{{x}}\: \\ $$$$\:\frac{{a}}{{x}}\:=\:\frac{{x}}{{b}}\:\Leftrightarrow\:{x}^{\mathrm{2}} \:=\:{ab}\:\Leftrightarrow\:{x}=\sqrt{{ab}} \\ $$…

Question-206729

Question Number 206729 by mr W last updated on 23/Apr/24 Answered by A5T last updated on 23/Apr/24 $$\frac{{sin}\mathrm{84}}{{AD}}=\frac{{sin}\mathrm{54}}{{AC}}\Rightarrow{AD}=\frac{{ACsin}\mathrm{84}}{{sin}\mathrm{54}}…\left({i}\right) \\ $$$$\frac{{sin}\left(?\right)}{{AD}}=\frac{{sin}\left(\mathrm{54}−?\right)}{{BD}={AC}}\Rightarrow{AD}=\frac{{ACsin}\left(?\right)}{{sin}\left(\mathrm{54}−?\right)}…\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow\frac{{sin}\mathrm{84}}{{sin}\mathrm{54}}=\frac{{sin}\left(?\right)}{{sin}\left(\mathrm{54}−?\right)}\Rightarrow?=\mathrm{30}° \\ $$ Commented…

Question-206615

Question Number 206615 by cortano21 last updated on 20/Apr/24 Answered by mr W last updated on 20/Apr/24 $$\frac{{R}−{r}}{{R}+{r}}=\mathrm{sin}\:\mathrm{22}.\mathrm{5}°=\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}{R}\right){R}=\left(\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\right){r} \\ $$$$\frac{{r}}{{R}}=\frac{\mathrm{2}−\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}{\mathrm{2}+\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}}\approx\mathrm{0}.\mathrm{4465} \\ $$ Terms…