Category: Geometry

Question Number 182066 by HeferH last updated on 03/Dec/22 Answered by a.lgnaoui last updated on 04/Dec/22 $$△\mathrm{ABC}$$$$\mathrm{ABcos}2\mathrm{x}+\mathrm{BCcos}\mathrm{x}=\mathrm{ADcos}3\mathrm{x}+\mathrm{CDcos}4\mathrm{x}$$$$\frac{\sin 2\mathrm{x}}{\mathrm{BC}}=\frac{\mathrm{sinx}}{\mathrm{AB}}\left(1\right)\frac{\sin 3\mathrm{x}}{\mathrm{CD}}=\frac{\sin 4\mathrm{x}}{\mathrm{AD}}\left(2\right)$$$$\mathrm{ABsin}2\mathrm{x}=\mathrm{BCsin}\mathrm{x}\mathrm{AD}\sin 3\mathrm{x}=\mathrm{CDsin}4\mathrm{x}$$$$\mathrm{AB}=\frac{\mathrm{BCsin}\:\mathrm{x}}{\mathrm{sin}\:\mathrm{2x}}\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{CD}=\frac{\mathrm{ADsin}\:\mathrm{3x}}{\mathrm{sin}\:\mathrm{4x}}…

Question Number 116466 by mr W last updated on 04/Oct/20 Commented by mr W last updated on 04/Oct/20 $$R=radiusofbigcircles$$$$r=radiusofsmallcircles$$$$let\lambda =\frac{r}{R}$$$$\mathrm{cos}\:\alpha=\frac{{R}}{{R}+{r}}=\frac{\mathrm{1}}{\mathrm{1}+\lambda}…

Question Number 181987 by Acem last updated on 03/Dec/22 Answered by Frix last updated on 03/Dec/22 $$\frac{\sin 45°}{BC}=\frac{\sin 105°}{AC}=\frac{\sin 30°}{AB})$$$$\Rightarrow \frac{BC}{AC}=\sqrt{3}-1$$$$AC+BC=14{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{get}$$$$AC=\frac{14\sqrt{3}}{3}\wedge BC=\frac{14(3-\sqrt{3})}{3}\Rightarrow AB=\frac{7\sqrt{2}(3-\sqrt{3})}{3}$$…

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Question Number 181918 by HeferH last updated on 02/Dec/22 Answered by Acem last updated on 02/Dec/22 Commented by Acem last updated on 02/Dec/22 $${Sir}\:{HeferH},\:{share}\:{with}\:{me}\:{to}\:{find}\:{x}\:{cause} \…