Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 74509 by ajfour last updated on 25/Nov/19

Commented by ajfour last updated on 25/Nov/19

Find the radius of circle. △ABC is equilateral with side unity.

$${Find}\:{the}\:{radius}\:{of}\:{circle}.\:\bigtriangleup{ABC} \\ $$$${is}\:{equilateral}\:{with}\:{side}\:{unity}. \\ $$

Commented by ajfour last updated on 25/Nov/19

Commented by ajfour last updated on 25/Nov/19

AT= r(√3) , sin α = (r/(1−r(√3))) cos (60°−α)= BF ((cos α+(√3)sin α)/2)=(1−r(√3))cos α ⇒ 1+(√3)tan α = 2−2r(√3) ⇒ ((r(√3))/(√((1−r(√3))^2 −r^2 ))) = 1−2r(√3) ⇒ 3r^2 =(1−2r(√3))^2 (1−2r(√3)+2r^2 ) ⇒ −3r^2 +1−2r(√3)+2r^2 +12r^2 −24(√3)r^3 +24r^4 −4(√3)r+24r^2 −8(√3)r^3 = 0 ⇒ 24r^4 −32(√3)r^3 +35r^2 −6(√3)r+1=0 possibly r ≈ 0.16739 (&1.5293)

$${AT}=\:{r}\sqrt{\mathrm{3}}\:,\:\:\mathrm{sin}\:\alpha\:=\:\frac{{r}}{\mathrm{1}−{r}\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cos}\:\left(\mathrm{60}°−\alpha\right)=\:{BF} \\ $$$$\frac{\mathrm{cos}\:\alpha+\sqrt{\mathrm{3}}\mathrm{sin}\:\alpha}{\mathrm{2}}=\left(\mathrm{1}−{r}\sqrt{\mathrm{3}}\right)\mathrm{cos}\:\alpha \\ $$$$\Rightarrow\:\:\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{tan}\:\alpha\:=\:\mathrm{2}−\mathrm{2}{r}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:\:\frac{{r}\sqrt{\mathrm{3}}}{\sqrt{\left(\mathrm{1}−{r}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} }}\:=\:\mathrm{1}−\mathrm{2}{r}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{3}{r}^{\mathrm{2}} =\left(\mathrm{1}−\mathrm{2}{r}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{r}\sqrt{\mathrm{3}}+\mathrm{2}{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\:−\mathrm{3}{r}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{r}\sqrt{\mathrm{3}}+\mathrm{2}{r}^{\mathrm{2}} +\mathrm{12}{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:−\mathrm{24}\sqrt{\mathrm{3}}{r}^{\mathrm{3}} +\mathrm{24}{r}^{\mathrm{4}} −\mathrm{4}\sqrt{\mathrm{3}}{r}+\mathrm{24}{r}^{\mathrm{2}} \\ $$$$\:\:\:\:\:−\mathrm{8}\sqrt{\mathrm{3}}{r}^{\mathrm{3}} \:=\:\mathrm{0}\:\:\:\Rightarrow \\ $$$$\:\mathrm{24}{r}^{\mathrm{4}} −\mathrm{32}\sqrt{\mathrm{3}}{r}^{\mathrm{3}} +\mathrm{35}{r}^{\mathrm{2}} −\mathrm{6}\sqrt{\mathrm{3}}{r}+\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:{possibly}\:\:\:{r}\:\approx\:\mathrm{0}.\mathrm{16739}\:\:\left(\&\mathrm{1}.\mathrm{5293}\right) \\ $$

Commented by ajfour last updated on 25/Nov/19

Terms of Service

Privacy Policy

Contact: info@tinkutara.com