Category: Geometry

Question Number 199728 by Mingma last updated on 08/Nov/23 Answered by AST last updated on 08/Nov/23 $$\frac{sin120°}{BC}=\frac{sin20°}{AB}\Rightarrow AB=\frac{2\sqrt{3}BCsin20°}{3}$$$$\frac{sin20°}{AB}=\frac{sin40°}{AC}\Rightarrow AC=2ABcos20°=\frac{2\sqrt{3}BCsin40}{3}$$$$\frac{sin\left(x\right)}{DC}=\frac{sinADC}{AC}=\frac{sin(140-x)}{AC}\Rightarrow \frac{AC}{DC}=\frac{sin(140-x)}{sin\left(x\right)}$$$$=\frac{sin40}{sin20}=2cos20=\frac{sin140cosx-sin\left(x\right)cos\left(140\right)}{sinx}$$$$={sin}\mathrm{40}{tan}\left({x}\right)−{cos}\left(\mathrm{90}+\mathrm{50}\right)={sin}\mathrm{40}{tanx}+{sin}\mathrm{50}…

Question Number 199672 by Mingma last updated on 07/Nov/23 Answered by mr W last updated on 07/Nov/23 Commented by mr W last updated on 07/Nov/23…

Question Number 199510 by universe last updated on 04/Nov/23 Answered by mr W last updated on 06/Nov/23 Commented by mr W last updated on 06/Nov/23…

Question Number 199413 by universe last updated on 03/Nov/23 Answered by ajfour last updated on 03/Nov/23 $$b\cos \alpha =R$$$${\{R\tan \alpha +\left(\frac{a}{b}\right)R\}}^2+{\left\{\left(\frac{a}{b}\right)R\tan \alpha \right\}}^2=R^2$$$$say\tan \alpha =t$$$$\Rightarrow\:{t}^{\mathrm{2}}…

Question Number 199358 by ajfour last updated on 01/Nov/23 Commented by ajfour last updated on 01/Nov/23 $$FindR.$$ Answered by mr W last updated…