Category: Geometry

Question Number 195896 by AROUNAMoussa last updated on 12/Aug/23 Answered by HeferH last updated on 13/Aug/23 Commented by HeferH last updated on 13/Aug/23 $$\mathrm{180}°−\mathrm{4}{x}\:+\:\mathrm{40}°+\mathrm{2}{x}\:=\:\mathrm{180}° \…

Question Number 195880 by aawb_247 last updated on 12/Aug/23 Commented by aawb_247 last updated on 12/Aug/23 $$anyonecanhelp..whatavalueofx?$$ Answered by mr W last updated…

Question Number 195740 by universe last updated on 09/Aug/23 Answered by Frix last updated on 09/Aug/23 $$\mathrm{Triangle}\Rightarrow a+b>c\wedge a+c>b\wedge b+c>a$$$$\mathrm{Let}b=(u-v)a\wedge c=(u+v)a$$$$\Rightarrow u>\frac{1}{2}\wedge -\frac{1}{2}<v<\frac{1}{2}$$$$f(u,v)=\frac{1}{2u}+\frac{u-v}{u+v+1}+\frac{u+v}{u-v+1}$$$$\frac{\mathrm{3}}{\mathrm{2}}\leqslant{f}\left({u},\:{v}\right)<\mathrm{2}…

Question Number 195344 by Mingma last updated on 31/Jul/23 Answered by kapoorshah last updated on 31/Jul/23 $$cos\alpha =cos\alpha$$$$\frac{{a}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{a}.\mathrm{6}}\:=\:\frac{{b}^{\mathrm{2}} \:+\:\mathrm{6}^{\mathrm{2}} \:−\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}.{b}.\mathrm{6}}…

Question Number 195287 by Mingma last updated on 29/Jul/23 Answered by MM42 last updated on 29/Jul/23 $$CD=DE\Rightarrow \mathrm{\angle }C2=\mathrm{\angle }E\mathrm{\&}\mathrm{\angle }D1+\mathrm{\angle }E=60$$$$\mathrm{\angle }C1=60-\mathrm{\angle }C2=60-\mathrm{\angle }E=\mathrm{\angle }D1$$$$\frac{CD}{Sin60}=\frac{2}{SinC1}\mathrm{\&}\frac{DE}{Sin120}=\frac{BE}{SinD1}$$$$\frac{2}{SinC1}=\frac{BE}{SinD1}\Rightarrow BE=2✓$$…