Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 177784 by mnjuly1970 last updated on 09/Oct/22

Answered by aleks041103 last updated on 09/Oct/22

Commented by aleks041103 last updated on 09/Oct/22

construct  H∈OO′, s.t. BH⊥OO′  ΔOO′A(∡O=90°)⇒∡O′+∡A=90°  ⇒α+β=90°  But AOB is a quarter circle  ⇒∡AO′B=90°  But  ∡OO′A+∡AO′B+∡BO′H=180°  ⇒α+∡BO′H=90°  ⇒∡BO′H=β  But by construction for ΔHO′B(∡H=90°)  ⇒∡HO′B+∡O′BH=90°  ⇒∡O′BH=α  ⇒ΔO′OA=ΔBHO′  since:  1. both are right Δ  2. ∡O′BH=∡OO′A  3. AO′=O′B=r  ⇒BH=OO′=(R/2)  ∡OHB=90° and OB=R  ⇒OH=(√(R^2 −((R/2))^2 ))=((√3)/2)R  ⇒O′H=OH−OO′=(((√3)−1)/2)R  ⇒O′B=(√(((((√3)−1)/2)R)^2 +((R/2))^2 ))=  =(R/2)(√(((√3)−1)^2 +1))=((√(5−2(√3)))/2)R=r  ⇒(r/R)=((√(5−2(√3)))/2)

$${construct} \\ $$$${H}\in{OO}',\:{s}.{t}.\:{BH}\bot{OO}' \\ $$$$\Delta{OO}'{A}\left(\measuredangle{O}=\mathrm{90}°\right)\Rightarrow\measuredangle{O}'+\measuredangle{A}=\mathrm{90}° \\ $$$$\Rightarrow\alpha+\beta=\mathrm{90}° \\ $$$${But}\:{AOB}\:{is}\:{a}\:{quarter}\:{circle} \\ $$$$\Rightarrow\measuredangle{AO}'{B}=\mathrm{90}° \\ $$$${But} \\ $$$$\measuredangle{OO}'{A}+\measuredangle{AO}'{B}+\measuredangle{BO}'{H}=\mathrm{180}° \\ $$$$\Rightarrow\alpha+\measuredangle{BO}'{H}=\mathrm{90}° \\ $$$$\Rightarrow\measuredangle{BO}'{H}=\beta \\ $$$${But}\:{by}\:{construction}\:{for}\:\Delta{HO}'{B}\left(\measuredangle{H}=\mathrm{90}°\right) \\ $$$$\Rightarrow\measuredangle{HO}'{B}+\measuredangle{O}'{BH}=\mathrm{90}° \\ $$$$\Rightarrow\measuredangle{O}'{BH}=\alpha \\ $$$$\Rightarrow\Delta{O}'{OA}=\Delta{BHO}' \\ $$$${since}: \\ $$$$\mathrm{1}.\:{both}\:{are}\:{right}\:\Delta \\ $$$$\mathrm{2}.\:\measuredangle{O}'{BH}=\measuredangle{OO}'{A} \\ $$$$\mathrm{3}.\:{AO}'={O}'{B}={r} \\ $$$$\Rightarrow{BH}={OO}'=\frac{{R}}{\mathrm{2}} \\ $$$$\measuredangle{OHB}=\mathrm{90}°\:{and}\:{OB}={R} \\ $$$$\Rightarrow{OH}=\sqrt{{R}^{\mathrm{2}} −\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{R} \\ $$$$\Rightarrow{O}'{H}={OH}−{OO}'=\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}{R} \\ $$$$\Rightarrow{O}'{B}=\sqrt{\left(\frac{\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{2}}{R}\right)^{\mathrm{2}} +\left(\frac{{R}}{\mathrm{2}}\right)^{\mathrm{2}} }= \\ $$$$=\frac{{R}}{\mathrm{2}}\sqrt{\left(\sqrt{\mathrm{3}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}=\frac{\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}}}{\mathrm{2}}{R}={r} \\ $$$$\Rightarrow\frac{{r}}{{R}}=\frac{\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{3}}}}{\mathrm{2}} \\ $$

Commented by Tawa11 last updated on 09/Oct/22

Great sir.

$$\mathrm{Great}\:\mathrm{sir}. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com