Category: Geometry

Question Number 65153 by Tawa1 last updated on 25/Jul/19 Commented by Tony Lin last updated on 25/Jul/19 Commented by Tawa1 last updated on 25/Jul/19 $$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}…

Question Number 65128 by Tawa1 last updated on 25/Jul/19 Answered by ajfour last updated on 25/Jul/19 Commented by ajfour last updated on 25/Jul/19 $${OP}\:=\mathrm{8} \\…

Question Number 64971 by Tawa1 last updated on 23/Jul/19 Commented by Tony Lin last updated on 23/Jul/19 $$\frac{\frac{{x}}{\mathrm{2}}}{{sin}\theta}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{sin}\left(\mathrm{90}°+\theta\right)}=\frac{{x}}{{sin}\left(\mathrm{90}°−\mathrm{2}\theta\right)} \\ $$$$\Rightarrow\frac{\frac{{x}}{\mathrm{2}}}{{sin}\theta}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{cos}\theta}=\frac{{x}}{{cos}\mathrm{2}\theta} \\ $$$${let}\:{cos}\theta={t} \\ $$$$\Rightarrow\frac{\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\frac{\frac{{x}}{\mathrm{2}}+\mathrm{1}}{{t}}=\frac{{x}}{\mathrm{2}{t}^{\mathrm{2}}…

Question Number 64909 by Tawa1 last updated on 23/Jul/19 Commented by MJS last updated on 23/Jul/19 $$\mathrm{the}\:\mathrm{picture}\:\mathrm{shows}\:\mathrm{2}\:\mathrm{circles},\:\mathrm{so}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{numbers}\:\mathrm{don}'\mathrm{t}\:\mathrm{fit}.\:\mathrm{obviously}\:\mathrm{the}\:\mathrm{right}\:\mathrm{must} \\ $$$$\mathrm{be}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{the}\:\mathrm{above}/\mathrm{below}\:\mathrm{numbers} \\ $$ Commented by…

Question Number 64892 by mathmax by abdo last updated on 22/Jul/19 Terms of Service Privacy Policy Contact: info@tinkutara.com