Category: Geometry

Question Number 176195 by adhigenz last updated on 14/Sep/22 Answered by behi834171 last updated on 15/Sep/22 $$BC=BP=PC=\frac{\sqrt{3}}{3},AB=AR=BR=\frac{2\sqrt{3}}{3}$$$${PQ}^2=\frac{1}{3}+1-2\times \frac{\sqrt{3}}{3}\times 1\times cos\left(150\right)=$$$$=\frac{1}{3}+1+2\times \frac{\sqrt{3}}{3}\times \frac{\sqrt{3}}{2}\Rightarrow PQ=\sqrt{\frac{7}{3}}$$$${cos}\measuredangle{CPQ}=\frac{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{7}}{\mathrm{3}}}\right)^{\mathrm{2}}…

Question Number 176100 by mnjuly1970 last updated on 12/Sep/22 $$$$$$Q:$$$$inA\stackrel{\mathrm{\Delta }}{BC}:A={90}^{\mathrm{o}}and,m_b^2+m_c^2=100.$$$$$$$$\:\:\:\:\:\:\:\:\:\:\:{find}\:\:{the}\:\:{value}\:{of}\:\:\:''\:\:\:{a}\:\:\:''\:. \…

Question Number 176076 by Agnibhoo98 last updated on 11/Sep/22 Commented by som(math1967) last updated on 11/Sep/22 $$\mathrm{\angle }\mathit{B}AC=90°$$ Terms of Service Privacy Policy Contact:…

Question Number 176014 by cherokeesay last updated on 11/Sep/22 Commented by cherokeesay last updated on 11/Sep/22 $$ThetriangleABCisright-angledatB.$$ Answered by cortano1 last updated on…

Question Number 175717 by ajfour last updated on 05/Sep/22 Answered by mr W last updated on 05/Sep/22 $${(R-r)}^2=r^2+{(\frac{R}{2}+r)}^2$$$${r}^{\mathrm{2}} +\mathrm{3}{Rr}−\frac{\mathrm{3}{R}^{\mathrm{2}} }{\mathrm{4}}=\mathrm{0} \…