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 45690 by ajfour last updated on 15/Oct/18

Commented by ajfour last updated on 15/Oct/18

Radius of circle is unity. Find b in terms of a.

$${Radius}\:{of}\:{circle}\:{is}\:{unity}. \\ $$$${Find}\:{b}\:{in}\:{terms}\:{of}\:{a}. \\ $$

Answered by MrW3 last updated on 15/Oct/18

Commented by ajfour last updated on 15/Oct/18

Thank you so much Sir; it is indeed a matter of faith. BEAUTIFUL !

$${Thank}\:{you}\:{so}\:{much}\:{Sir};\:{it}\:{is} \\ $$$${indeed}\:{a}\:{matter}\:{of}\:{faith}. \\ $$$$\mathcal{BEAUTIFUL}\:! \\ $$

Commented by MrW3 last updated on 15/Oct/18

thank you too sir!

$${thank}\:{you}\:{too}\:{sir}! \\ $$

Commented by MrW3 last updated on 15/Oct/18

tan α=((√((2R)^2 −a^2 ))/a)=((√(4R^2 −a^2 ))/a) γ=2α PT=R tan γ=R tan (2α)=R((2(√(4R^2 −a^2 )))/(a(1−((4R^2 −a^2 )/a^2 ))))=((aR(√(4R^2 −a^2 )))/(a^2 −2R^2 )) tan β=((PT)/(2R))=((a(√(4R^2 −a^2 )))/(2(a^2 −2R^2 ))) cos β=((2(a^2 −2R^2 ))/(√(a^2 (4R^2 −a^2 )+4(a^2 −2R^2 )^2 ))) cos β=((2(a^2 −2R^2 ))/(√(3a^4 +16R^4 −12a^2 R^2 ))) cos β=((2(a^2 −2R^2 ))/(√(3(a^2 −2R^2 )^2 +4R^4 ))) b=2R cos β ⇒b=((4R(a^2 −2R^2 ))/(√(3(a^2 −2R^2 )^2 +4R^4 ))) for R=1: ⇒b=((4(a^2 −2))/(√(3(a^2 −2)^2 +4)))

$$\mathrm{tan}\:\alpha=\frac{\sqrt{\left(\mathrm{2}{R}\right)^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}}=\frac{\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}} \\ $$$$\gamma=\mathrm{2}\alpha \\ $$$${PT}={R}\:\mathrm{tan}\:\gamma={R}\:\mathrm{tan}\:\left(\mathrm{2}\alpha\right)={R}\frac{\mathrm{2}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}\left(\mathrm{1}−\frac{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}=\frac{{aR}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\beta=\frac{{PT}}{\mathrm{2}{R}}=\frac{{a}\sqrt{\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\sqrt{{a}^{\mathrm{2}} \left(\mathrm{4}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\sqrt{\mathrm{3}{a}^{\mathrm{4}} +\mathrm{16}{R}^{\mathrm{4}} −\mathrm{12}{a}^{\mathrm{2}} {R}^{\mathrm{2}} }} \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\sqrt{\mathrm{3}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{4}} }} \\ $$$${b}=\mathrm{2}{R}\:\mathrm{cos}\:\beta \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}{R}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)}{\sqrt{\mathrm{3}\left({a}^{\mathrm{2}} −\mathrm{2}{R}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{4}{R}^{\mathrm{4}} }} \\ $$$${for}\:{R}=\mathrm{1}: \\ $$$$\Rightarrow{b}=\frac{\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{2}\right)}{\sqrt{\mathrm{3}\left({a}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} +\mathrm{4}}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com