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Category: Geometry

Question-163292

Question Number 163292 by amin96 last updated on 05/Jan/22 Answered by mr W last updated on 05/Jan/22 $${say}\:{CD}={AD}=\mathrm{1} \\ $$$${AC}=\mathrm{2}\:\mathrm{cos}\:\mathrm{10}° \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{20}}=\frac{{AC}}{\mathrm{sin}\:\left(\mathrm{30}+\mathrm{20}\right)} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{2}\:\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{20}}{\mathrm{sin}\:\mathrm{50}} \\…

Question-163143

Question Number 163143 by amin96 last updated on 04/Jan/22 Commented by HongKing last updated on 04/Jan/22 $$\mathrm{r}^{\mathrm{2}} \:=\:\mathrm{r}\:+\:\mathrm{2021}\:+\:\mathrm{2021}^{\mathrm{2}} \\ $$$$\mathrm{r}^{\mathrm{2}} \:-\:\mathrm{r}\:-\:\mathrm{2021}^{\mathrm{2}} \:-\:\mathrm{2021}\:=\:\mathrm{0} \\ $$$$\mathrm{r}\:=\:-\:\mathrm{2021}\:;\:\mathrm{2022}\:\Rightarrow\:\mathrm{r}\:=\:\mathrm{2022}\:\checkmark \\…

Question-97557

Question Number 97557 by Power last updated on 08/Jun/20 Answered by mr W last updated on 08/Jun/20 $$\angle{CBE}=\angle{EBD}=\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{\mathrm{2}}−\alpha\right) \\ $$$$\Rightarrow\mathrm{2}\theta=\frac{\pi}{\mathrm{2}}−\alpha \\ $$$${BD}=\mathrm{2}{r}\:\mathrm{sin}\:\alpha \\ $$$${EB}=\frac{\mathrm{2}{r}}{\mathrm{cos}\:\theta} \\…