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 158543 by ajfour last updated on 06/Nov/21

Commented by ajfour last updated on 06/Nov/21

The quadrilateral is a square.

$${The}\:{quadrilateral}\:{is}\:{a}\:{square}. \\ $$

Commented by mr W last updated on 06/Nov/21

Commented by mr W last updated on 06/Nov/21

tan θ=a ϕ=(π/4)+(θ/2) Radius big circle=a+(a/(sin ϕ))=(a/(sin θ cos θ)) 1+(1/(sin ((π/4)+(θ/2))))=(1/(sin θ cos θ)) ⇒θ≈35.0446° ⇒a≈0.7014

$$\mathrm{tan}\:\theta={a} \\ $$$$\varphi=\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}} \\ $$$${Radius}\:{big}\:{circle}={a}+\frac{{a}}{\mathrm{sin}\:\varphi}=\frac{{a}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{4}}+\frac{\theta}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\theta\approx\mathrm{35}.\mathrm{0446}° \\ $$$$\Rightarrow{a}\approx\mathrm{0}.\mathrm{7014} \\ $$

Commented by Tawa11 last updated on 06/Nov/21

Weldone sir

$$\mathrm{Weldone}\:\mathrm{sir} \\ $$

Answered by ajfour last updated on 06/Nov/21

let a=tan θ 1+a^2 =R (R−a)^2 =a^2 +b^2 a^2 +a^4 ={(a^2 +b)sin θ+b}^2 ⇒ (1+a^2 −a)^2 =a^2 +b^2 a^2 +a^4 ={((a(a^2 +b)+b(√(1+a^2 )))/( (√(1+a^2 ))))}^2 a(1+a^2 )=a^3 + (a+(√(1+a^2 )))(1−a)(√(1+a^2 )) ⇒ a((√(1+a^2 ))−a)=(1−a)(√(1+a^2 )) ⇒ (2a−1)(√(1+a^2 ))=a^2 ⇒ (2−(1/a))(√(1+(1/a^2 )))=1 ⇒ a≈0.70136 (>(1/2))

$${let}\:\:{a}=\mathrm{tan}\:\theta \\ $$$$\mathrm{1}+{a}^{\mathrm{2}} ={R} \\ $$$$\left({R}−{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{a}^{\mathrm{4}} =\left\{\left({a}^{\mathrm{2}} +{b}\right)\mathrm{sin}\:\theta+{b}\right\}^{\mathrm{2}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}+{a}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{a}^{\mathrm{4}} =\left\{\frac{{a}\left({a}^{\mathrm{2}} +{b}\right)+{b}\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}{\:\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }}\right\}^{\mathrm{2}} \\ $$$${a}\left(\mathrm{1}+{a}^{\mathrm{2}} \right)={a}^{\mathrm{3}} + \\ $$$$\:\:\:\:\:\:\:\left({a}+\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }\right)\left(\mathrm{1}−{a}\right)\sqrt{\mathrm{1}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$${a}\left(\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }−{a}\right)=\left(\mathrm{1}−{a}\right)\sqrt{\mathrm{1}+{a}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\left(\mathrm{2}{a}−\mathrm{1}\right)\sqrt{\mathrm{1}+{a}^{\mathrm{2}} }={a}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left(\mathrm{2}−\frac{\mathrm{1}}{{a}}\right)\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}=\mathrm{1} \\ $$$$\Rightarrow\:\:{a}\approx\mathrm{0}.\mathrm{70136}\:\:\left(>\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com