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 54022 by ajfour last updated on 28/Jan/19

Commented by ajfour last updated on 28/Jan/19

$$Givenaandb,find\theta andRinterms$$$$ofaandb.$$

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jan/19

Commented by ajfour last updated on 28/Jan/19

Commented by ajfour last updated on 28/Jan/19

$$AE=AD+DE$$$$\frac{b}{\sin \theta }=a\cot \frac{\theta }{2}+\sqrt{{(a+b)}^2-a^2}$$$$\frac{b}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}-\frac{a\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}=\sqrt{b(b+2a)}$$$$\frac{b-a(1+\cos \theta )}{\sin \theta }=\sqrt{b(b+2a)}$$$$\Rightarrow b^2+a^2{(1+\cos \theta )}^2-2ab(1+\cos \theta )$$$$=b(b+2a)\sin {}^2\theta$$$$b^2+a^2+2a^2\cos \theta +a^2\cos {}^2\theta -2ab$$$$-2ab\cos \theta =b^2+2ab-b^2\cos {}^2\theta -2ab\cos {}^2\theta$$$$let\cos \theta =s$$$$\Rightarrow {(a+b)}^2{\mathit{s}}^2+2a(a-b)\mathit{s}-a(4b-a)=0$$$$\mathit{s}=\frac{a(a-b)+\sqrt{a^2{(a-b)}^2+a{(a+b)}^2(4b-a)}}{{(a+b)}^2}$$$$aandRbotharerootsof$$$${(x+b)}^2{\mathit{s}}^2+2x(x-b)\mathit{s}-x(4b-x)=0$$$$\Rightarrow {\mathit{x}}^2{(1+\mathit{s})}^2+2b\mathit{x}({\mathit{s}}^2-\mathit{s}-2)+b^2{\mathit{s}}^2=0$$$$\Rightarrow \mathit{a}\mathit{R}=\frac{b^2{\mathit{s}}^2}{{(1+\mathit{s})}^2}$$$$hence\mathit{R}=\frac{b^2{\mathit{s}}^2}{a{(1+\mathit{s})}^2}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com