Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 30159 by naka3546 last updated on 17/Feb/18

Commented by mrW2 last updated on 18/Feb/18

$$sir,pleasecheckyourquestion.$$$$Ithinkthegivenconfigurationleads$$$$tononormalsolution.$$$$IfBD=5cm,thenACmustbebetween$$$$7.5and9cmtomakeasolution$$$$possible.E.g.ifyousetAC=8cm,$$$${you}^{\prime }llgetDC=\frac{40}{3}cm.$$

Answered by mrW2 last updated on 18/Feb/18

Commented by mrW2 last updated on 18/Feb/18

$$let\lambda =\frac{a}{b}$$$$\frac{a}{\sin 3\theta }=\frac{CB}{\sin 4\theta }...\left(i\right)$$$$\frac{b}{\sin \theta }=\frac{CB}{\sin 2\theta }...\left(ii\right)$$$$\left(i\right)/\left(ii\right):$$$$\frac{a}{\sin 3\theta }\times \frac{\sin \theta }{b}=\frac{\sin 2\theta }{\sin 4\theta }=\frac{1}{2\cos 2\theta }$$$$\Rightarrow \frac{\lambda \sin \theta }{\sin 3\theta }=\frac{1}{2\cos 2\theta }$$$$2\lambda \sin \theta \cos 2\theta =\sin 3\theta =\sin \theta \cos 2\theta +\cos \theta \sin 2\theta$$$$(2\lambda -1)\sin \theta \cos 2\theta =\cos \theta \sin 2\theta$$$$(2\lambda -1)\tan \theta =\tan 2\theta =\frac{2\tan \theta }{1-{\tan }^2\theta }$$$$2\lambda -1=\frac{2}{1-{\tan }^2\theta }$$$${\tan }^2\theta =1-\frac{2}{2\lambda -1}>0\Rightarrow \lambda >\frac{3}{2}$$$$\Rightarrow \tan \theta =\sqrt{1-\frac{2}{2\lambda -1}}$$$$3\theta +4\theta <180°$$$$\Rightarrow \theta <\frac{180°}{7}\approx 2$$$${\tan }^2\theta =1-\frac{2}{2\lambda -1}<{\tan }^2\left(\frac{180°}{7}\right)$$$$\Rightarrow \lambda <\frac{1}{2}+\frac{1}{1-{\tan }^2\frac{180°}{7}}=1.8$$$$i.e.tomakeasolutionpossible,$$$$1.5<\lambda =\frac{a}{b}<1.8$$$$forb=5cm:$$$$a>1.5\times 5=7.5cm$$$$a<1.8\times 5=9cm$$$$$$$${let}^{\prime }ssayb=8cm,\Rightarrow \lambda =\frac{8}{5}=1.6$$$$\Rightarrow \tan \theta =\sqrt{1-\frac{2}{2\times 1.6-1}}=\frac{1}{\sqrt{11}}\Rightarrow \theta =16.8°$$$$\Rightarrow \cos 2\theta =2{\cos }^2\theta -1=\frac{2}{1+{\tan }^2\theta }-1$$$$\Rightarrow \cos 2\theta =\frac{2}{1+\frac{1}{11}}-1=\frac{2\times 11}{12}-1=\frac{5}{6}$$$$\frac{DC}{\sin 4\theta }=\frac{a}{\sin 2\theta }$$$$\Rightarrow DC=\frac{a\sin 4\theta }{\sin 2\theta }=2a\cos 2\theta =2\times 8\times \frac{5}{6}=\frac{40}{3}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com