Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 28911 by ajfour last updated on 01/Feb/18

Commented by ajfour last updated on 01/Feb/18

$$Findareaofcolouredtriangle$$$$andradiusofcircleintermsof$$$$\mathit{a},\mathit{\theta },and\mathit{\varphi }.Assume\theta ,\varphi ⩽\frac{\pi }{4}.$$$$Estimatemaximumvalueof$$$$radiusintermsof\mathit{a}.$$

Answered by mrW2 last updated on 01/Feb/18

Commented by mrW2 last updated on 01/Feb/18

$$BF=\frac{a}{\cos \theta }$$$$DF=a(1-\tan \theta )$$$$DE=a(1-\tan \varphi )$$$$\tan \phi =\frac{DE}{DF}=\frac{1-\tan \varphi }{1-\tan \theta }$$$$\Rightarrow \phi ={\tan }^{-1}\left(\frac{1-\tan \varphi }{1-\tan \theta }\right)$$$$\alpha =90-(\theta +\varphi )$$$$\beta =180-(90-\theta )-\phi =90-(\phi -\theta )$$$$\frac{r}{\tan \frac{\alpha }{2}}+\frac{r}{\tan \frac{\beta }{2}}=\frac{a}{\cos \theta }$$$$\frac{1}{\tan \frac{\alpha }{2}}=\frac{1+\cos \alpha }{\sin \alpha }=\frac{1+\sin (\theta +\varphi )}{\cos (\theta +\varphi )}$$$$\frac{1}{\tan \frac{\beta }{2}}=\frac{1+\cos \beta }{\sin \beta }=\frac{1+\sin (\phi -\theta )}{\cos (\phi -\theta )}$$$$\Rightarrow r[\frac{1+\sin (\varphi +\theta )}{\cos (\varphi +\theta )}+\frac{1+\sin (\phi -\theta )}{\cos (\phi -\theta )}]=\frac{a}{\cos \theta }$$$$\Rightarrow r=\frac{a\cos (\varphi +\theta )\cos (\phi -\theta )}{\cos \theta [\cos (\varphi +\theta )+\cos (\phi -\theta )+\sin (\varphi +\theta )\cos (\phi -\theta )+\cos (\varphi +\theta )\sin (\phi -\theta )]}$$$$\Rightarrow r=\frac{a\cos (\varphi +\theta )\cos (\phi -\theta )}{\cos \theta [\cos (\varphi +\theta )+\cos (\phi -\theta )+\sin (\varphi +\phi )]}$$$$$$$$A_{\mathrm{\Delta }}=\frac{1}{2}\times \frac{a}{\cos \theta }\times \frac{a}{\cos \varphi }\times \sin \alpha =\frac{a^2}{2}\times \frac{\cos (\theta +\varphi )}{\cos \theta \cos \varphi }$$$$\Rightarrow A_{\mathrm{\Delta }}=\frac{a^2}{2}\times \frac{\cos (\theta +\varphi )}{\cos \theta \cos \varphi }=\frac{a^2}{2}(1-\tan \theta \tan \varphi )$$$$$$$$Formax.ofradiusr:$$$$\varphi =\theta$$$$\phi =\frac{\pi }{4}$$$$\Rightarrow r=\frac{a\cos \left(2\theta \right)\cos (\frac{\pi }{4}-\theta )}{\cos \theta [\cos \left(2\theta \right)+\cos (\frac{\pi }{4}-\theta )+\sin (\frac{\pi }{4}+\theta )]}$$$$\Rightarrow r=\frac{a\cos \left(2\theta \right)\cos (\frac{\pi }{4}-\theta )}{\cos \theta [\cos \left(2\theta \right)+2\cos (\frac{\pi }{4}-\theta )]}$$$$\Rightarrow r=\frac{a\cos \left(2\theta \right)(\cos \theta +\sin \theta )\times \frac{1}{\sqrt{2}}}{\cos \theta [\cos \left(2\theta \right)+\sqrt{2}(\cos \theta +\sin \theta )]}$$$$\Rightarrow r=\frac{a\cos \left(2\theta \right)(\cos \theta +\sin \theta )}{\cos \theta [\sqrt{2}\cos \left(2\theta \right)+2(\cos \theta +\sin \theta )]}$$$$\Rightarrow r=\frac{a\cos \left(2\theta \right)}{\cos \theta [\sqrt{2}(\cos \theta -\sin \theta )+2]}$$$$f\left(\theta \right)=\frac{\cos \left(2\theta \right)}{\cos \theta [\sqrt{2}(\cos \theta -\sin \theta )+2]}$$$$f\left(\theta \right)=\frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta [\sqrt{2}(1-\tan \theta )+\frac{2}{\cos \theta }]}$$$$f\left(\theta \right)=\frac{1-{\tan }^2\theta }{\sqrt{2}(1-\tan \theta )+2\sqrt{1+{\tan }^2\theta }}$$$$f\left(\theta \right)=\frac{1-x^2}{\sqrt{2}(1-x)+2\sqrt{1+x^2}}withx=\tan \theta$$$$$$$$f{\left(\theta \right)}_{max}\approx 0.3032at\theta \approx 9.3°$$$$\Rightarrow r_{max}\approx 0.3032a$$

Commented by ajfour last updated on 01/Feb/18

$$Area(△BEF)=\frac{1}{2}\left(BE\right)\left(BF\right)\sin \alpha$$$$=\frac{{\mathit{a}}^2\cos (\mathit{\theta }+\mathit{\varphi })}{2\cos \mathit{\theta }\cos \mathit{\varphi }}=\frac{\mathit{r}}{2}(\mathit{B}\mathit{E}+\mathit{B}\mathit{F}+\mathit{E}\mathit{F})$$$$\Rightarrow \mathit{r}(\frac{\mathit{a}}{\cos \mathit{\varphi }}+\frac{\mathit{a}}{\cos \mathit{\theta }}+\mathit{E}\mathit{F})=\frac{{\mathit{a}}^2\cos (\mathit{\theta }+\mathit{\varphi })}{\cos \mathit{\theta }\cos \mathit{\varphi }}$$$$\mathit{E}\mathit{F}=\mathit{a}\sqrt{{(1-\tan \mathit{\theta })}^2+{(1-\tan \mathit{\varphi })}^2}$$$$\mathit{r}=\mathit{a}\left[\frac{\cos (\mathit{\theta }+\mathit{\varphi })}{\cos \mathit{\theta }+\cos \mathit{\varphi }+\sqrt{2}\left(\sqrt{\cos {}^2\mathit{\varphi }\sin {}^2(\mathit{\theta }-\mathit{\pi }/4)+\cos {}^2\mathit{\theta }\sin {}^2(\mathit{\varphi }-\mathit{\pi }/4)}\right)}\right]$$$$expressionfor\mathit{r},if\mathit{\theta }=\mathit{\varphi },$$$$\mathit{r}=\mathit{a}\left[\frac{\cos 2\mathit{\theta }}{2\cos \mathit{\theta }+2\cos \mathit{\theta }\mid \sin (\mathit{\theta }-\mathit{\pi }/4)\mid }\right]$$$$=\frac{\mathit{a}\cos 2\mathit{\theta }}{2\cos \mathit{\theta }[1+\sin (\frac{\mathit{\pi }}{4}-\mathit{\theta })]}$$$$=\frac{\mathit{a}(1-{\mathit{x}}^2)}{2(\sqrt{1+x^2}+\frac{1}{\sqrt{2}}-\frac{\mathit{x}}{\sqrt{2}})}(ifwelet\mathit{x}=\tan \mathit{\theta })$$$$\mathit{r}\left(x\right)=\frac{\mathit{a}}{\sqrt{2}}\left(\frac{1-{\mathit{x}}^2}{\sqrt{2(1+x^2)}+1-\mathit{x}}\right)$$$$\frac{dr}{dx}=0\Rightarrow x=?\left(pleasehelp\right)$$$$\frac{\mathit{d}\mathit{r}}{\mathit{d}\mathit{x}}=0leadstotheequation$$$${(x-1)}^4(1+x^2)=2x^2{(3+x^2)}^2$$$$finditsrootsthatbelongto(0,1).$$

Commented by ajfour last updated on 01/Feb/18

Thank you mrW2 Sir !

Commented by mrW2 last updated on 01/Feb/18

Thank you sir! I could form my formula to the same expression as yours. I think it is not possible to solve x analytically. I found my solution through graph. x=0.165

Commented by ajfour last updated on 01/Feb/18

Which graph app sir?

Commented by mrW2 last updated on 01/Feb/18

One can use e.g. Geogebra.

Commented by mrW2 last updated on 01/Feb/18

Commented by ajfour last updated on 01/Feb/18

$$thanksSir.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com